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A sequence is an ordered list of infinitely many
numbers that may or may not have a pattern.
Sequences
A sequence is an ordered list of infinitely many
numbers that may or may not have a pattern.
Example A.
1, 3, 5, 7, 9,… is the sequence of odd numbers.
Sequences
A sequence is an ordered list of infinitely many
numbers that may or may not have a pattern.
Example A.
1, 3, 5, 7, 9,… is the sequence of odd numbers.
1, 4, 9, 16, 25, …is the sequence of square numbers.
Sequences
A sequence is an ordered list of infinitely many
numbers that may or may not have a pattern.
Example A.
1, 3, 5, 7, 9,… is the sequence of odd numbers.
1, 4, 9, 16, 25, …is the sequence of square numbers.
5, –2, , e2, –110, …is a sequence without an
obvious pattern.
Sequences
A sequence is an ordered list of infinitely many
numbers that may or may not have a pattern.
Example A.
1, 3, 5, 7, 9,… is the sequence of odd numbers.
1, 4, 9, 16, 25, …is the sequence of square numbers.
5, –2, , e2, –110, …is a sequence without an
obvious pattern.
Sequences
One way to describe a sequence is to give a formula
a(n) for its terms and list the sequence as {a(n)}n=1.
∞
A sequence is an ordered list of infinitely many
numbers that may or may not have a pattern.
Example A.
1, 3, 5, 7, 9,… is the sequence of odd numbers.
1, 4, 9, 16, 25, …is the sequence of square numbers.
5, –2, , e2, –110, …is a sequence without an
obvious pattern.
Sequences
One way to describe a sequence is to give a formula
a(n) for its terms and list the sequence as {a(n)}n=1.
We also write a(n) as an.
∞
A sequence is an ordered list of infinitely many
numbers that may or may not have a pattern.
Example A.
1, 3, 5, 7, 9,… is the sequence of odd numbers.
1, 4, 9, 16, 25, …is the sequence of square numbers.
5, –2, , e2, –110, …is a sequence without an
obvious pattern.
Sequences
One way to describe a sequence is to give a formula
a(n) for its terms and list the sequence as {a(n)}n=1.
We also write a(n) as an.
∞
Example B.
a. The sequence {an = 3n + 1}n=1 = {4, 7, 10, …}
with a1 = 4, a2 = 7, a3 = 10, …
A sequence is an ordered list of infinitely many
numbers that may or may not have a pattern.
Example A.
1, 3, 5, 7, 9,… is the sequence of odd numbers.
1, 4, 9, 16, 25, …is the sequence of square numbers.
5, –2, , e2, –110, …is a sequence without an
obvious pattern.
Sequences
One way to describe a sequence is to give a formula
a(n) for its terms and list the sequence as {a(n)}n=1.
We also write a(n) as an.
∞
Example B.
a. The sequence {an = 3n + 1}n=1 = {4, 7, 10, …}
with a1 = 4, a2 = 7, a3 = 10, …
From here on, it’s assumed
n → ∞ at the top
A sequence can have multiple representations.
b. The sequence 0, 0, 0, 0,.. may be listed as
{a(n) = 0}n=1 or as {sin(nπ)}n=1.
Example C. a. The sequence {(–1)n(2n – 1)}n=1
gives the alternating sequence –1, 3, –5, 7, –9, …
∞
b. Find a formula for, , , , ...with k = 0, 1, 2...
–4
9
2
3
6
27
–8
81
The formula 2(k + 1) gives the numerators,
3k+1 gives the denominators, and (–1)k switches
the ± signs so {(–1)k2(k+1)/ 3k+1}k=0 is one solution.∞
Sequences
The multiple factor (–1)n or (–1)n+1 alternate the signs
of the terms.
Sequences whose terms have alternating ± signs
are called alternating sequences.
Sequences
The sequence 1, –1/2, 1/3, –1/4.. → 0,
the sequence –2/1, –3/2, –4/3, –5/4,.. → –1,
and cos(1/2), cos(1/3), cos(1/4), .. → 1.
Convergent Sequences
Sequences
We say the sequence {an} converges (CG) or that
it’s a convergent (CG) sequence if lim an = L is finite.n∞
The sequence 1, –1/2, 1/3, –1/4.. → 0,
the sequence –2/1, –3/2, –4/3, –5/4,.. → –1,
and cos(1/2), cos(1/3), cos(1/4), .. → 1.
Convergent Sequences
Sequences
We say the sequence {an} converges (CG) or that
it’s a convergent (CG) sequence if lim an = L is finite.
All the above sequences are CG-sequences.
n∞
The sequence 1, –1/2, 1/3, –1/4.. → 0,
the sequence –2/1, –3/2, –4/3, –5/4,.. → –1,
and cos(1/2), cos(1/3), cos(1/4), .. → 1.
Convergent Sequences
Sequences
We say the sequence {an} converges (CG) or that
it’s a convergent (CG) sequence if lim an = L is finite.
All the above sequences are CG-sequences.
By lim an = L we mean that:n∞
n∞
The sequence 1, –1/2, 1/3, –1/4.. → 0,
the sequence –2/1, –3/2, –4/3, –5/4,.. → –1,
and cos(1/2), cos(1/3), cos(1/4), .. → 1.
Convergent Sequences
Sequences
We say the sequence {an} converges (CG) or that
it’s a convergent (CG) sequence if lim an = L is finite.
All the above sequences are CG-sequences.
L
No matter how small
n∞
By lim an = L we mean that: for every ϵ > 0,n∞
The sequence 1, –1/2, 1/3, –1/4.. → 0,
the sequence –2/1, –3/2, –4/3, –5/4,.. → –1,
and cos(1/2), cos(1/3), cos(1/4), .. → 1.
Convergent Sequences
Sequences
We say the sequence {an} converges (CG) or that
it’s a convergent (CG) sequence if lim an = L is finite.
All the above sequences are CG-sequences.
L
No matter how small
n∞
By lim an = L we mean that: for every ϵ > 0,
all except finitely many an’s, are inside the interval
(L – ϵ, L + ϵ),
n∞
The sequence 1, –1/2, 1/3, –1/4.. → 0,
the sequence –2/1, –3/2, –4/3, –5/4,.. → –1,
and cos(1/2), cos(1/3), cos(1/4), .. → 1.
Convergent Sequences
Sequences
We say the sequence {an} converges (CG) or that
it’s a convergent (CG) sequence if lim an = L is finite.
All the above sequences are CG-sequences.
LL–ϵ L+ϵ
No matter how small of an interval is
roped off around L,
n∞
By lim an = L we mean that: for every ϵ > 0,
all except finitely many an’s, are inside the interval
(L – ϵ, L + ϵ),
n∞
The sequence 1, –1/2, 1/3, –1/4.. → 0,
the sequence –2/1, –3/2, –4/3, –5/4,.. → –1,
and cos(1/2), cos(1/3), cos(1/4), .. → 1.
Convergent Sequences
Sequences
We say the sequence {an} converges (CG) or that
it’s a convergent (CG) sequence if lim an = L is finite.
All the above sequences are CG-sequences.
No matter how small of an interval is
roped off around L, all except finitely many
an’s, are inside the interval.
n∞
By lim an = L we mean that: for every ϵ > 0,
all except finitely many an’s, are inside the interval
(L – ϵ, L + ϵ),
n∞
The sequence 1, –1/2, 1/3, –1/4.. → 0,
the sequence –2/1, –3/2, –4/3, –5/4,.. → –1,
and cos(1/2), cos(1/3), cos(1/4), .. → 1.
Convergent Sequences
Lan’s L–ϵ L+ϵ an’s
Sequences
We say the sequence {an} converges (CG) or that
it’s a convergent (CG) sequence if lim an = L is finite.
All the above sequences are CG-sequences.
By lim an = L we mean that: for every ϵ > 0,
all except finitely many an’s, are inside the interval
(L – ϵ, L + ϵ),
Lan’s L–ϵ L+ϵ an’s
n∞
No matter how small of an interval is
roped off around L, all except finitely many
an’s, are inside the interval.
Finitely many an’s are outside
n∞
The sequence 1, –1/2, 1/3, –1/4.. → 0,
the sequence –2/1, –3/2, –4/3, –5/4,.. → –1,
and cos(1/2), cos(1/3), cos(1/4), .. → 1.
Convergent Sequences
Sequences
We say the sequence {an} converges (CG) or that
it’s a convergent (CG) sequence if lim an = L is finite.
All the above sequences are CG-sequences.
The sequence 1, –1/2, 1/3, –1/4.. → 0,
the sequence –2/1, –3/2, –4/3, –5/4,.. → –1,
and cos(1/2), cos(1/3), cos(1/4), .. → 1.
n∞
By lim an = L we mean that: for every ϵ > 0,
all except finitely many an’s, are inside the interval
(L – ϵ, L + ϵ),
which is equivalent to
saying that:
“for sufficiently large n,
we have l an – L l < ϵ”.
n∞
Convergent Sequences
Lan’s L–ϵ L+ϵ an’s
No matter how small of an interval is
roped off around L, all except finitely many
an’s, are inside the interval.
Finitely many an’s are outside
Sequences
Convergent Sequences
Sequences
The sequence of numbers 1/2, 2/3, 3/4, 4/5,..
are defined by {an= n/(n+1)}n=1 and they correspond
to the points on the graph of y = x/(x + 1) as shown.
∞
y = 1
y = x/(x + 1)(1,1/2)
(2,2/3)
(3,3/4) (4,4/5)
Convergent Sequences
Sequences
The sequence of numbers 1/2, 2/3, 3/4, 4/5,..
are defined by {an= n/(n+1)}n=1 and they correspond
to the points on the graph of y = x/(x + 1) as shown.
∞
y = 1
y = x/(x + 1)(1,1/2)
(2,2/3)
(3,3/4) (4,4/5)Since lim x/(x + 1) = 1 and
these points are on its graph,
we have lim an= 1 as n→∞.
x→∞
n∞
Convergent Sequences
Sequences
The sequence of numbers 1/2, 2/3, 3/4, 4/5,..
are defined by {an= n/(n+1)}n=1 and they correspond
to the points on the graph of y = x/(x + 1) as shown.
∞
y = 1
y = x/(x + 1)(1,1/2)
(2,2/3)
(3,3/4) (4,4/5)Since lim x/(x + 1) = 1 and
these points are on its graph,
we have lim an= 1 as n→∞.
x→∞
n∞
(Function-Sequence Convergent Theorem)
Given a sequence {fn} defined by y = f(x),
if lim f(x) = L is finite, then lim fn = L.x→∞
Convergent Sequences
Sequences
The sequence of numbers 1/2, 2/3, 3/4, 4/5,..
are defined by {an= n/(n+1)}n=1 and they correspond
to the points on the graph of y = x/(x + 1) as shown.
∞
y = 1
y = x/(x + 1)(1,1/2)
(2,2/3)
(3,3/4) (4,4/5)Since lim x/(x + 1) = 1 and
these points are on its graph,
we have lim an= 1 as n→∞.
x→∞
n∞
(Function-Sequence Convergent Theorem)
Given a sequence {fn} defined by y = f(x),
if lim f(x) = L is finite, then lim fn = L.
In the case that f(x) is differentiable, we may use the
L’Hopital Rule to see if f(x) converges; if the defining
function f(x) converges then the sequence converges.
x→∞
Convergent Sequences
Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ .
n∞
= 0 by L’Hopital Rule
Hence {an = n1/n } is a convergent sequence
and lim an = 1.
n
Let’s find the limit of the defining function x1/x.
Example E. a. Show that {an = Ln(n)/n1/2} →0.
Lim Ln(x)/x1/2 lim = lim 2x–1/2 = 0.x–1
½ x–1/2
Hence {an = Ln(n)/n1/2} is a CG sequence.
b. Show that {bn = n20/en} is a CG sequence.
L‘Hospital
Rule
Since [x20](21) = 0, so by applying the L‘Hospital Rule
21 times we’ve x20/ex → 0 so that {n20/en} converges.
→
Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
n
Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
n
Let’s find the limit of the defining function x1/x.
Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x)
n
Let’s find the limit of the defining function x1/x.
Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x
n
Let’s find the limit of the defining function x1/x.
Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x
= 0 by L’Hopital Rule
n
Let’s find the limit of the defining function x1/x.
Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ .
= 0 by L’Hopital Rule
n
Let’s find the limit of the defining function x1/x.
Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ .
n∞
= 0 by L’Hopital Rule
Hence {an = n1/n } is a convergent sequence
and lim an = 1.
n
Let’s find the limit of the defining function x1/x.
Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ .
n∞
= 0 by L’Hopital Rule
Hence {an = n1/n } is a convergent sequence
and lim an = 1.
n
Let’s find the limit of the defining function x1/x.
Example E. a. Show that {an = Ln(n)/n1/2} →0.
Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ .
n∞
= 0 by L’Hopital Rule
Hence {an = n1/n } is a convergent sequence
and lim an = 1.
n
Let’s find the limit of the defining function x1/x.
Example E. a. Show that {an = Ln(n)/n1/2} →0.
Lim Ln(x)/x1/2 lim x–1
½ x–1/2
L‘Hospital
Rule
→
Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ .
n∞
= 0 by L’Hopital Rule
Hence {an = n1/n } is a convergent sequence
and lim an = 1.
n
Let’s find the limit of the defining function x1/x.
Example E. a. Show that {an = Ln(n)/n1/2} →0.
Lim Ln(x)/x1/2 lim = lim 2x–1/2 = 0.x–1
½ x–1/2
L‘Hospital
Rule
→
Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ .
n∞
= 0 by L’Hopital Rule
Hence {an = n1/n } is a convergent sequence
and lim an = 1.
n
Let’s find the limit of the defining function x1/x.
Example E. a. Show that {an = Ln(n)/n1/2} →0.
Lim Ln(x)/x1/2 lim = lim 2x–1/2 = 0.x–1
½ x–1/2
Hence {an = Ln(n)/n1/2} is a CG sequence.
L‘Hospital
Rule
→
Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ .
n∞
= 0 by L’Hopital Rule
Hence {an = n1/n } is a convergent sequence
and lim an = 1.
n
Let’s find the limit of the defining function x1/x.
Example E. a. Show that {an = Ln(n)/n1/2} →0.
Lim Ln(x)/x1/2 lim = lim 2x–1/2 = 0.x–1
½ x–1/2
Hence {an = Ln(n)/n1/2} is a CG sequence.
b. Show that {bn = n20/en} is a CG sequence.
L‘Hospital
Rule
→
Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ .
n∞
= 0 by L’Hopital Rule
Hence {an = n1/n } is a convergent sequence
and lim an = 1.
n
Let’s find the limit of the defining function x1/x.
Example E. a. Show that {an = Ln(n)/n1/2} →0.
Lim Ln(x)/x1/2 lim = lim 2x–1/2 = 0.x–1
½ x–1/2
Hence {an = Ln(n)/n1/2} is a CG sequence.
b. Show that {bn = n20/en} is a CG sequence.
L‘Hospital
Rule
Applying the L‘Hospital Rule and differentiate 21 times
we have [x20](21) = 0, so that {n20/en} → 0 converges.
→
Sequences
As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞,
but at different paces as shown in example E.
Sequences
As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞,
but at different paces as shown in example E.
L’Hopital Rule gives us the following general results,
in short, as x→∞:
Sequences
As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞,
but at different paces as shown in example E.
L’Hopital Rule gives us the following general results,
in short, as x→∞:
* Ln(x) goes to ∞ “slower” than xp
Sequences
As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞,
but at different paces as shown in example E.
L’Hopital Rule gives us the following general results,
in short, as x→∞:
* Ln(x) goes to ∞ “slower” than xp and
* xp goes to ∞ “slower” than ex.
Sequences
As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞,
but at different paces as shown in example E.
L’Hopital Rule gives us the following general results,
in short, as x→∞:
* Ln(x) goes to ∞ “slower” than xp and
* xp goes to ∞ “slower” than ex.
(Sin(x) and cos(x) are bounded between –1 and 1.)
Sequences
As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞,
but at different paces as shown in example E.
L’Hopital Rule gives us the following general results,
in short, as x→∞:
* Ln(x) goes to ∞ “slower” than xp and
* xp goes to ∞ “slower” than ex.
(Sin(x) and cos(x) are bounded between –1 and 1.)
Here is a useful theorem for justifying convergence.
Sequences
As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞,
but at different paces as shown in example E.
L’Hopital Rule gives us the following general results,
in short, as x→∞:
* Ln(x) goes to ∞ “slower” than xp and
* xp goes to ∞ “slower” than ex.
(Sin(x) and cos(x) are bounded between –1 and 1.)
n∞ n∞
(The Sandwich Theorem)
Let 0 ≤ an ≤ bn for all but finitely many n’s
and that lim bn = 0, then lim an = 0.
Here is a useful theorem for justifying convergence.
Sequences
As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞,
but at different paces as shown in example E.
L’Hopital Rule gives us the following general results,
in short, as x→∞:
* Ln(x) goes to ∞ “slower” than xp and
* xp goes to ∞ “slower” than ex.
(Sin(x) and cos(x) are bounded between –1 and 1.)
n∞ n∞
(The Sandwich Theorem)
Let 0 ≤ an ≤ bn for all but finitely many n’s
and that lim bn = 0, then lim an = 0.
The Sandwich Theorem offers another method in
justifying convergence besides by the L’Hopital Rule.
Here is a useful theorem for justifying convergence.
Sequences
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
Compare bn = {1011/n} to an = {10n/n!}.
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
We will show that bn > an (> 0) for sufficiently large n,
Compare bn = {1011/n} to an = {10n/n!}.
and since bn0 we must have an0.
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
We will show that bn > an (> 0) for sufficiently large n,
Compare bn = {1011/n} to an = {10n/n!}.
and since bn0 we must have an0.
Specifically we claim that bn > an for 10 < n = 11,12,..
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
We will show that bn > an (> 0) for sufficiently large n,
For n > 10, note the following blocks in their products
10n
n! = 10 * 10….. 10*10*10*10…10
n * (n–1)…11*10* 9 * 8…..1
Compare bn = {1011/n} to an = {10n/n!}.
and since bn0 we must have an0.
Specifically we claim that bn > an for 10 < n = 11,12,.. :
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
an =
Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
We will show that bn > an (> 0) for sufficiently large n,
For n > 10, note the following blocks in their products
10n
n! = 10 * 10….. 10*10*10*10…10
n * (n–1)…11*10* 9 * 8…..1
1
Compare bn = {1011/n} to an = {10n/n!}.
and since bn0 we must have an0.
Specifically we claim that bn > an for 10 < n = 11,12,.. :
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
an =
Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
We will show that bn > an (> 0) for sufficiently large n,
For n > 10, note the following blocks in their products
10n
n! = 10 * 10….. 10*10*10*10…10
n * (n–1)…11*10* 9 * 8…..1
10101
Compare bn = {1011/n} to an = {10n/n!}.
and since bn0 we must have an0.
Specifically we claim that bn > an for 10 < n = 11,12,.. :
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
an =
Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
We will show that bn > an (> 0) for sufficiently large n,
For n > 10, note the following blocks in their products
10n
n! = 10 * 10….. 10*10*10*10…10
n * (n–1)…11*10* 9 * 8…..1
10101
< 1011
n
Compare bn = {1011/n} to an = {10n/n!}.
and since bn0 we must have an0.
Specifically we claim that bn > an for 10 < n = 11,12,.. :
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
an = = bn
Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
We will show that bn > an (> 0) for sufficiently large n,
For n > 10, note the following blocks in their products
10n
n! = 10 * 10….. 10*10*10*10…10
n * (n–1)…11*10* 9 * 8…..1
10101
< 1011
n
n∞
Compare bn = {1011/n} to an = {10n/n!}.
and since bn0 we must have an0.
Specifically we claim that bn > an for 10 < n = 11,12,.. :
0
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
an = = bn
Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
We will show that bn > an (> 0) for sufficiently large n,
For n > 10, note the following blocks in their products
10n
n! = 10 * 10….. 10*10*10*10…10
n * (n–1)…11*10* 9 * 8…..1
10101
< 1011
n
So for large n, bn = 1011/n > an = 10n/n!.
n∞
Compare bn = {1011/n} to an = {10n/n!}.
and since bn0 we must have an0.
Specifically we claim that bn > an for 10 < n = 11,12,.. :
0
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
an = = bn
Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
We will show that bn > an (> 0) for sufficiently large n,
For n > 10, note the following blocks in their products
10n
n! = 10 * 10….. 10*10*10*10…10
n * (n–1)…11*10* 9 * 8…..1
10101
< 1011
n
So for large n, bn = 1011/n > an = 10n/n!. Since bn0,
therefore lim10n/n! = 0 by the sandwich theorem.
n∞
Compare bn = {1011/n} to an = {10n/n!}.
and since bn0 we must have an0.
Specifically we claim that bn > an for 10 < n = 11,12,.. :
0
n∞
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
an = = bn
Sequences
Properties of CG Sequences
Sequences
Properties of CG Sequences
Let lim an = L and lim bn = K be CG sequences,
1. If c is a constant, {can} converges and lim can = cL.
Sequences
Properties of CG Sequences
Let lim an = L and lim bn = K be CG sequences,
1. If c is a constant, {can} converges and lim can = cL.
2. {an± bn}, {an bn} and {an/bn} converge
Sequences
Properties of CG Sequences
Let lim an = L and lim bn = K be CG sequences,
1. If c is a constant, {can} converges and lim can = cL.
2. {an± bn}, {an bn} and {an/bn} converge with
lim (an ± bn) = L ± K, lim anbn = (lim an)(lim bn) = LK,
and lim (an / bn) = (lim an) / (lim bn) = L / K if K ≠ 0.
Sequences
Properties of CG Sequences
Let lim an = L and lim bn = K be CG sequences,
1. If c is a constant, {can} converges and lim can = cL.
2. {an± bn}, {an bn} and {an/bn} converge with
lim (an ± bn) = L ± K, lim anbn = (lim an)(lim bn) = LK,
and lim (an / bn) = (lim an) / (lim bn) = L / K if K ≠ 0.
Example G. Given that {an = 2 + 1/n} → 2,
{bn = 3cos(1/n)} → 3 as n →∞, then
Sequences
Properties of CG Sequences
Let lim an = L and lim bn = K be CG sequences,
1. If c is a constant, {can} converges and lim can = cL.
2. {an± bn}, {an bn} and {an/bn} converge with
lim (an ± bn) = L ± K, lim anbn = (lim an)(lim bn) = LK,
and lim (an / bn) = (lim an) / (lim bn) = L / K if K ≠ 0.
Example G. Given that {an = 2 + 1/n} → 2,
{bn = 3cos(1/n)} → 3 as n →∞, then
* {10an = 10(2 + 1/n)} → 10(2) = 20
Sequences
Properties of CG Sequences
Let lim an = L and lim bn = K be CG sequences,
1. If c is a constant, {can} converges and lim can = cL.
2. {an± bn}, {an bn} and {an/bn} converge with
lim (an ± bn) = L ± K, lim anbn = (lim an)(lim bn) = LK,
and lim (an / bn) = (lim an) / (lim bn) = L / K if K ≠ 0.
Example G. Given that {an = 2 + 1/n} → 2,
{bn = 3cos(1/n)} → 3 as n →∞, then
* {10an = 10(2 + 1/n)} → 10(2) = 20
* {an * bn = (2 + 1/n) * 3cos(1/n)}
Sequences
Properties of CG Sequences
Let lim an = L and lim bn = K be CG sequences,
1. If c is a constant, {can} converges and lim can = cL.
2. {an± bn}, {an bn} and {an/bn} converge with
lim (an ± bn) = L ± K, lim anbn = (lim an)(lim bn) = LK,
and lim (an / bn) = (lim an) / (lim bn) = L / K if K ≠ 0.
Example G. Given that {an = 2 + 1/n} → 2,
{bn = 3cos(1/n)} → 3 as n →∞, then
* {10an = 10(2 + 1/n)} → 10(2) = 20
* {an * bn = (2 + 1/n) * 3cos(1/n)} → (2)(3) = 6
Sequences
Properties of CG Sequences
Let lim an = L and lim bn = K be CG sequences,
1. If c is a constant, {can} converges and lim can = cL.
2. {an± bn}, {an bn} and {an/bn} converge with
lim (an ± bn) = L ± K, lim anbn = (lim an)(lim bn) = LK,
and lim (an / bn) = (lim an) / (lim bn) = L / K if K ≠ 0.
Example G. Given that {an = 2 + 1/n} → 2,
{bn = 3cos(1/n)} → 3 as n →∞, then
* {10an = 10(2 + 1/n)} → 10(2) = 20
* {an * bn = (2 + 1/n) * 3cos(1/n)} → (2)(3) = 6
* {an / bn = (2 + 1/n) / 3cos(1/n)} → 2/3.
Sequences
A non-convergent sequence is a called
a divergent (DG) sequence.
Divergent Sequences
Sequences
A non-convergent sequence is a called
a divergent (DG) sequence. In other words,
a sequence {an} diverges (DG) if lim an is UDF.n∞
Divergent Sequences
Sequences
A non-convergent sequence is a called
a divergent (DG) sequence. In other words,
a sequence {an} diverges (DG) if lim an is UDF.n∞
Divergent Sequences
A nice sequence {fn} might be defined by a badly
behaved function f(x).
Sequences
A non-convergent sequence is a called
a divergent (DG) sequence. In other words,
a sequence {an} diverges (DG) if lim an is UDF.n∞
Divergent Sequences
A nice sequence {fn} might be defined by a badly
behaved function f(x).
Example G.
The sequence 0, 0, 0,.. is defined by {fn= n*sin(n)}
which match the x-intercepts of
y = x*sin(x).
Sequences
A non-convergent sequence is a called
a divergent (DG) sequence. In other words,
a sequence {an} diverges (DG) if lim an is UDF.n∞
Divergent Sequences
A nice sequence {fn} might be defined by a badly
behaved function f(x).
Example G.
(1,0) (2,0) (3,0) (4,0)
y = x*sin(x)
The sequence 0, 0, 0,.. is defined by {fn= n*sin(n)}
which match the x-intercepts of
y = x*sin(x).
Sequences
A non-convergent sequence is a called
a divergent (DG) sequence. In other words,
a sequence {an} diverges (DG) if lim an is UDF.n∞
Divergent Sequences
A nice sequence {fn} might be defined by a badly
behaved function f(x).
Example G.
x→∞
(1,0) (2,0) (3,0) (4,0)
y = x*sin(x)
The sequence 0, 0, 0,.. is defined by {fn= n*sin(n)}
which match the x-intercepts of
y = x*sin(x).
but lim x*sin(x) is undefined.
Lim fn = 0 hence
{fn= n*sin(n)} converges,
Sequences
The convergence of the sequence does not imply the
convergence of the function because the sequence
corresponds to discrete points which contain a lot
less information than the graph itself.
Sequences
The convergence of the sequence does not imply the
convergence of the function because the sequence
corresponds to discrete points which contain a lot
less information than the graph itself.
Hence unlike the case for convergent sequences,
a divergent defining function f(x) is not enough to
justify that the sequence {fn} diverges.
Sequences
The convergence of the sequence does not imply the
convergence of the function because the sequence
corresponds to discrete points which contain a lot
less information than the graph itself.
Hence unlike the case for convergent sequences,
a divergent defining function f(x) is not enough to
justify that the sequence {fn} diverges.
We note the following two types of divergences,
the ones that go to ∞ and the ones that jump around
Sequences
The convergence of the sequence does not imply the
convergence of the function because the sequence
corresponds to discrete points which contain a lot
less information than the graph itself.
Hence unlike the case for convergent sequences,
a divergent defining function f(x) is not enough to
justify that the sequence {fn} diverges.
We note the following two types of divergences,
the ones that go to ∞ and the ones that jump around
* To-the-Infinity Divergence
Sequences
The convergence of the sequence does not imply the
convergence of the function because the sequence
corresponds to discrete points which contain a lot
less information than the graph itself.
Hence unlike the case for convergent sequences,
a divergent defining function f(x) is not enough to
justify that the sequence {fn} diverges.
We note the following two types of divergences,
the ones that go to ∞ and the ones that jump around
* To-the-Infinity Divergence
The function y = x → ∞ as x →∞,
so the sequence 1, 2, 3.. → +∞ as n →∞.
Sequences
The convergence of the sequence does not imply the
convergence of the function because the sequence
corresponds to discrete points which contain a lot
less information than the graph itself.
Hence unlike the case for convergent sequences,
a divergent defining function f(x) is not enough to
justify that the sequence {fn} diverges.
We note the following two types of divergences,
the ones that go to ∞ and the ones that jump around
* To-the-Infinity Divergence
The function y = x → ∞ as x →∞,
so the sequence 1, 2, 3.. → +∞ as n →∞.
The function y = –x2 → –∞ as x →∞,
so the sequence –1, –4, –9.. → –∞ as n →∞.
Sequences
(Divergence →±∞ )
If the defining function f(x) of fn goes to ±∞ as x →∞,
then lim fn → ±∞ as n →∞ and {fn} diverges.
Sequences
(Divergence →±∞ )
If the defining function f(x) of fn goes to ±∞ as x →∞,
then lim fn → ±∞ as n →∞ and {fn} diverges.
Specifically, by lim fn = ∞
we mean that for any
number C, we have that
fn > C for all except finitely
many fn’s as shown.
Sequences
(Divergence →±∞ )
If the defining function f(x) of fn goes to ±∞ as x →∞,
then lim fn → ±∞ as n →∞ and {fn} diverges.
Specifically, by lim fn = ∞
we mean that for any
number C, we have that
fn > C for all except finitely
many fn’s as shown.
C
(lim fn = ∞)
Given any number C,
Sequences
(Divergence →±∞ )
If the defining function f(x) of fn goes to ±∞ as x →∞,
then lim fn → ±∞ as n →∞ and {fn} diverges.
Specifically, by lim fn = ∞
we mean that for any
number C, we have that
fn > C for all except finitely
many fn’s as shown.
C
Given any number C,
all except finitely many fn’s, are > C.
(lim fn = ∞)
Sequences
(Divergence →±∞ )
If the defining function f(x) of fn goes to ±∞ as x →∞,
then lim fn → ±∞ as n →∞ and {fn} diverges.
Specifically, by lim fn = ∞
we mean that for any
number C, we have that
fn > C for all except finitely
many fn’s as shown.
C
Given any number C,
all except finitely many fn’s, are > C.
(lim fn = ∞)
a few fn’s < C
Sequences
(Divergence →±∞ )
If the defining function f(x) of fn goes to ±∞ as x →∞,
then lim fn → ±∞ as n →∞ and {fn} diverges.
We know the behavior of the terms of
a DG sequence that goes to →±∞. They blow up.
Specifically, by lim fn = ∞
we mean that for any
number C, we have that
fn > C for all except finitely
many fn’s as shown.
Ca few fn’s < C
Given any number C,
all except finitely many fn’s, are > C.
(lim fn = ∞)
Sequences
(Divergence →±∞ )
If the defining function f(x) of fn goes to ±∞ as x →∞,
then lim fn → ±∞ as n →∞ and {fn} diverges.
We know the behavior of the terms of
a DG sequence that goes to →±∞. They blow up.
Specifically, by lim fn = ∞
we mean that for any
number C, we have that
fn > C for all except finitely
many fn’s as shown.
Ca few fn’s < C
Given any number C,
all except finitely many fn’s, are > C.
(lim fn = ∞)
* Chaotic Divergence
Sequences
(Divergence →±∞ )
If the defining function f(x) of fn goes to ±∞ as x →∞,
then lim fn → ±∞ as n →∞ and {fn} diverges.
We know the behavior of the terms of
a DG sequence that goes to →±∞. They blow up.
Specifically, by lim fn = ∞
we mean that for any
number C, we have that
fn > C for all except finitely
many fn’s as shown.
Ca few fn’s < C
Given any number C,
all except finitely many fn’s, are > C.
(lim fn = ∞)
The sequence {sin(n)} = {sin(1)≈0.84, sin(2)≈0.90,
sin(3)≈0.14, ..} diverges because the terms jump
around between –1 and 1 without approaching a limit.
* Chaotic Divergence
Sequences
Note that for any number C between –1 and 1,
there are infinitely many sin(n)’s that are less than C
and there are also infinitely many sin(n)’s that are
more than C.
Sequences
Note that for any number C between –1 and 1,
there are infinitely many sin(n)’s that are less than C
and there are also infinitely many sin(n)’s that are
more than C. So unlike the case where lim fn = ∞,
we can’t predict the behavior of {sin(n)} as n →∞.
Sequences
Note that for any number C between –1 and 1,
there are infinitely many sin(n)’s that are less than C
and there are also infinitely many sin(n)’s that are
more than C. So unlike the case where lim fn = ∞,
we can’t predict the behavior of {sin(n)} as n →∞.
Hence {sin(n)} diverges chaotically.
Sequences
Note that for any number C between –1 and 1,
there are infinitely many sin(n)’s that are less than C
and there are also infinitely many sin(n)’s that are
more than C. So unlike the case where lim fn = ∞,
we can’t predict the behavior of {sin(n)} as n →∞.
Hence {sin(n)} diverges chaotically.
The oscillating behavior for chaotic divergence is
often caused by the periodic factors of sine or cosine
which equal to 0 periodically.
Sequences
Note that for any number C between –1 and 1,
there are infinitely many sin(n)’s that are less than C
and there are also infinitely many sin(n)’s that are
more than C. So unlike the case where lim fn = ∞,
we can’t predict the behavior of {sin(n)} as n →∞.
Hence {sin(n)} diverges chaotically.
The oscillating behavior for chaotic divergence is
often caused by the periodic factors of sine or cosine
which equal to 0 periodically.
Algebraic functions, Ln(x), ex, or xp (p > 0) are not
periodic so sequences defined by them converge,
or diverge to ±∞ (not chaotically).
For example,
{gn= n*sin(n/2)} = {1, 0, –3, 0, 5, 0, –7..}
does not converge so lim x*sin(x/2) must be UDF.
Sequences
Summary:
* a sequence {fn} inherits the “good” behaviors
such as lim f(x) = L or ±∞ of its defining function f(x).
* a function f(x) inherits the “bad” behavior of {fn}
so if lim f(n) doesn't exist then lim f(x) doesn't exist
For example, lim x/(x + 1) = 1 hence lim n/(n + 1) = 1
x→∞ n∞
* a CG sequence {fn} might be defined by
a non-convergent function f(x) as x →∞.
For example, the CG sequence 0, 0, 0.. is defined by
{fn= n*sin(n)} but y = x*sin(x) does not converge.

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20 sequences x

  • 1. A sequence is an ordered list of infinitely many numbers that may or may not have a pattern. Sequences
  • 2. A sequence is an ordered list of infinitely many numbers that may or may not have a pattern. Example A. 1, 3, 5, 7, 9,… is the sequence of odd numbers. Sequences
  • 3. A sequence is an ordered list of infinitely many numbers that may or may not have a pattern. Example A. 1, 3, 5, 7, 9,… is the sequence of odd numbers. 1, 4, 9, 16, 25, …is the sequence of square numbers. Sequences
  • 4. A sequence is an ordered list of infinitely many numbers that may or may not have a pattern. Example A. 1, 3, 5, 7, 9,… is the sequence of odd numbers. 1, 4, 9, 16, 25, …is the sequence of square numbers. 5, –2, , e2, –110, …is a sequence without an obvious pattern. Sequences
  • 5. A sequence is an ordered list of infinitely many numbers that may or may not have a pattern. Example A. 1, 3, 5, 7, 9,… is the sequence of odd numbers. 1, 4, 9, 16, 25, …is the sequence of square numbers. 5, –2, , e2, –110, …is a sequence without an obvious pattern. Sequences One way to describe a sequence is to give a formula a(n) for its terms and list the sequence as {a(n)}n=1. ∞
  • 6. A sequence is an ordered list of infinitely many numbers that may or may not have a pattern. Example A. 1, 3, 5, 7, 9,… is the sequence of odd numbers. 1, 4, 9, 16, 25, …is the sequence of square numbers. 5, –2, , e2, –110, …is a sequence without an obvious pattern. Sequences One way to describe a sequence is to give a formula a(n) for its terms and list the sequence as {a(n)}n=1. We also write a(n) as an. ∞
  • 7. A sequence is an ordered list of infinitely many numbers that may or may not have a pattern. Example A. 1, 3, 5, 7, 9,… is the sequence of odd numbers. 1, 4, 9, 16, 25, …is the sequence of square numbers. 5, –2, , e2, –110, …is a sequence without an obvious pattern. Sequences One way to describe a sequence is to give a formula a(n) for its terms and list the sequence as {a(n)}n=1. We also write a(n) as an. ∞ Example B. a. The sequence {an = 3n + 1}n=1 = {4, 7, 10, …} with a1 = 4, a2 = 7, a3 = 10, …
  • 8. A sequence is an ordered list of infinitely many numbers that may or may not have a pattern. Example A. 1, 3, 5, 7, 9,… is the sequence of odd numbers. 1, 4, 9, 16, 25, …is the sequence of square numbers. 5, –2, , e2, –110, …is a sequence without an obvious pattern. Sequences One way to describe a sequence is to give a formula a(n) for its terms and list the sequence as {a(n)}n=1. We also write a(n) as an. ∞ Example B. a. The sequence {an = 3n + 1}n=1 = {4, 7, 10, …} with a1 = 4, a2 = 7, a3 = 10, … From here on, it’s assumed n → ∞ at the top
  • 9. A sequence can have multiple representations. b. The sequence 0, 0, 0, 0,.. may be listed as {a(n) = 0}n=1 or as {sin(nπ)}n=1. Example C. a. The sequence {(–1)n(2n – 1)}n=1 gives the alternating sequence –1, 3, –5, 7, –9, … ∞ b. Find a formula for, , , , ...with k = 0, 1, 2... –4 9 2 3 6 27 –8 81 The formula 2(k + 1) gives the numerators, 3k+1 gives the denominators, and (–1)k switches the ± signs so {(–1)k2(k+1)/ 3k+1}k=0 is one solution.∞ Sequences The multiple factor (–1)n or (–1)n+1 alternate the signs of the terms. Sequences whose terms have alternating ± signs are called alternating sequences.
  • 10. Sequences The sequence 1, –1/2, 1/3, –1/4.. → 0, the sequence –2/1, –3/2, –4/3, –5/4,.. → –1, and cos(1/2), cos(1/3), cos(1/4), .. → 1. Convergent Sequences
  • 11. Sequences We say the sequence {an} converges (CG) or that it’s a convergent (CG) sequence if lim an = L is finite.n∞ The sequence 1, –1/2, 1/3, –1/4.. → 0, the sequence –2/1, –3/2, –4/3, –5/4,.. → –1, and cos(1/2), cos(1/3), cos(1/4), .. → 1. Convergent Sequences
  • 12. Sequences We say the sequence {an} converges (CG) or that it’s a convergent (CG) sequence if lim an = L is finite. All the above sequences are CG-sequences. n∞ The sequence 1, –1/2, 1/3, –1/4.. → 0, the sequence –2/1, –3/2, –4/3, –5/4,.. → –1, and cos(1/2), cos(1/3), cos(1/4), .. → 1. Convergent Sequences
  • 13. Sequences We say the sequence {an} converges (CG) or that it’s a convergent (CG) sequence if lim an = L is finite. All the above sequences are CG-sequences. By lim an = L we mean that:n∞ n∞ The sequence 1, –1/2, 1/3, –1/4.. → 0, the sequence –2/1, –3/2, –4/3, –5/4,.. → –1, and cos(1/2), cos(1/3), cos(1/4), .. → 1. Convergent Sequences
  • 14. Sequences We say the sequence {an} converges (CG) or that it’s a convergent (CG) sequence if lim an = L is finite. All the above sequences are CG-sequences. L No matter how small n∞ By lim an = L we mean that: for every ϵ > 0,n∞ The sequence 1, –1/2, 1/3, –1/4.. → 0, the sequence –2/1, –3/2, –4/3, –5/4,.. → –1, and cos(1/2), cos(1/3), cos(1/4), .. → 1. Convergent Sequences
  • 15. Sequences We say the sequence {an} converges (CG) or that it’s a convergent (CG) sequence if lim an = L is finite. All the above sequences are CG-sequences. L No matter how small n∞ By lim an = L we mean that: for every ϵ > 0, all except finitely many an’s, are inside the interval (L – ϵ, L + ϵ), n∞ The sequence 1, –1/2, 1/3, –1/4.. → 0, the sequence –2/1, –3/2, –4/3, –5/4,.. → –1, and cos(1/2), cos(1/3), cos(1/4), .. → 1. Convergent Sequences
  • 16. Sequences We say the sequence {an} converges (CG) or that it’s a convergent (CG) sequence if lim an = L is finite. All the above sequences are CG-sequences. LL–ϵ L+ϵ No matter how small of an interval is roped off around L, n∞ By lim an = L we mean that: for every ϵ > 0, all except finitely many an’s, are inside the interval (L – ϵ, L + ϵ), n∞ The sequence 1, –1/2, 1/3, –1/4.. → 0, the sequence –2/1, –3/2, –4/3, –5/4,.. → –1, and cos(1/2), cos(1/3), cos(1/4), .. → 1. Convergent Sequences
  • 17. Sequences We say the sequence {an} converges (CG) or that it’s a convergent (CG) sequence if lim an = L is finite. All the above sequences are CG-sequences. No matter how small of an interval is roped off around L, all except finitely many an’s, are inside the interval. n∞ By lim an = L we mean that: for every ϵ > 0, all except finitely many an’s, are inside the interval (L – ϵ, L + ϵ), n∞ The sequence 1, –1/2, 1/3, –1/4.. → 0, the sequence –2/1, –3/2, –4/3, –5/4,.. → –1, and cos(1/2), cos(1/3), cos(1/4), .. → 1. Convergent Sequences Lan’s L–ϵ L+ϵ an’s
  • 18. Sequences We say the sequence {an} converges (CG) or that it’s a convergent (CG) sequence if lim an = L is finite. All the above sequences are CG-sequences. By lim an = L we mean that: for every ϵ > 0, all except finitely many an’s, are inside the interval (L – ϵ, L + ϵ), Lan’s L–ϵ L+ϵ an’s n∞ No matter how small of an interval is roped off around L, all except finitely many an’s, are inside the interval. Finitely many an’s are outside n∞ The sequence 1, –1/2, 1/3, –1/4.. → 0, the sequence –2/1, –3/2, –4/3, –5/4,.. → –1, and cos(1/2), cos(1/3), cos(1/4), .. → 1. Convergent Sequences
  • 19. Sequences We say the sequence {an} converges (CG) or that it’s a convergent (CG) sequence if lim an = L is finite. All the above sequences are CG-sequences. The sequence 1, –1/2, 1/3, –1/4.. → 0, the sequence –2/1, –3/2, –4/3, –5/4,.. → –1, and cos(1/2), cos(1/3), cos(1/4), .. → 1. n∞ By lim an = L we mean that: for every ϵ > 0, all except finitely many an’s, are inside the interval (L – ϵ, L + ϵ), which is equivalent to saying that: “for sufficiently large n, we have l an – L l < ϵ”. n∞ Convergent Sequences Lan’s L–ϵ L+ϵ an’s No matter how small of an interval is roped off around L, all except finitely many an’s, are inside the interval. Finitely many an’s are outside
  • 21. Sequences The sequence of numbers 1/2, 2/3, 3/4, 4/5,.. are defined by {an= n/(n+1)}n=1 and they correspond to the points on the graph of y = x/(x + 1) as shown. ∞ y = 1 y = x/(x + 1)(1,1/2) (2,2/3) (3,3/4) (4,4/5) Convergent Sequences
  • 22. Sequences The sequence of numbers 1/2, 2/3, 3/4, 4/5,.. are defined by {an= n/(n+1)}n=1 and they correspond to the points on the graph of y = x/(x + 1) as shown. ∞ y = 1 y = x/(x + 1)(1,1/2) (2,2/3) (3,3/4) (4,4/5)Since lim x/(x + 1) = 1 and these points are on its graph, we have lim an= 1 as n→∞. x→∞ n∞ Convergent Sequences
  • 23. Sequences The sequence of numbers 1/2, 2/3, 3/4, 4/5,.. are defined by {an= n/(n+1)}n=1 and they correspond to the points on the graph of y = x/(x + 1) as shown. ∞ y = 1 y = x/(x + 1)(1,1/2) (2,2/3) (3,3/4) (4,4/5)Since lim x/(x + 1) = 1 and these points are on its graph, we have lim an= 1 as n→∞. x→∞ n∞ (Function-Sequence Convergent Theorem) Given a sequence {fn} defined by y = f(x), if lim f(x) = L is finite, then lim fn = L.x→∞ Convergent Sequences
  • 24. Sequences The sequence of numbers 1/2, 2/3, 3/4, 4/5,.. are defined by {an= n/(n+1)}n=1 and they correspond to the points on the graph of y = x/(x + 1) as shown. ∞ y = 1 y = x/(x + 1)(1,1/2) (2,2/3) (3,3/4) (4,4/5)Since lim x/(x + 1) = 1 and these points are on its graph, we have lim an= 1 as n→∞. x→∞ n∞ (Function-Sequence Convergent Theorem) Given a sequence {fn} defined by y = f(x), if lim f(x) = L is finite, then lim fn = L. In the case that f(x) is differentiable, we may use the L’Hopital Rule to see if f(x) converges; if the defining function f(x) converges then the sequence converges. x→∞ Convergent Sequences
  • 25. Sequences Example D. Is {an = n1/n = √n } a CG sequence? Find lim an if it’s a CG sequence. Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ . n∞ = 0 by L’Hopital Rule Hence {an = n1/n } is a convergent sequence and lim an = 1. n Let’s find the limit of the defining function x1/x. Example E. a. Show that {an = Ln(n)/n1/2} →0. Lim Ln(x)/x1/2 lim = lim 2x–1/2 = 0.x–1 ½ x–1/2 Hence {an = Ln(n)/n1/2} is a CG sequence. b. Show that {bn = n20/en} is a CG sequence. L‘Hospital Rule Since [x20](21) = 0, so by applying the L‘Hospital Rule 21 times we’ve x20/ex → 0 so that {n20/en} converges. →
  • 26. Sequences Example D. Is {an = n1/n = √n } a CG sequence? Find lim an if it’s a CG sequence. n
  • 27. Sequences Example D. Is {an = n1/n = √n } a CG sequence? Find lim an if it’s a CG sequence. n Let’s find the limit of the defining function x1/x.
  • 28. Sequences Example D. Is {an = n1/n = √n } a CG sequence? Find lim an if it’s a CG sequence. Lim x1/x = lim eLn(x)(1/x) n Let’s find the limit of the defining function x1/x.
  • 29. Sequences Example D. Is {an = n1/n = √n } a CG sequence? Find lim an if it’s a CG sequence. Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x n Let’s find the limit of the defining function x1/x.
  • 30. Sequences Example D. Is {an = n1/n = √n } a CG sequence? Find lim an if it’s a CG sequence. Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = 0 by L’Hopital Rule n Let’s find the limit of the defining function x1/x.
  • 31. Sequences Example D. Is {an = n1/n = √n } a CG sequence? Find lim an if it’s a CG sequence. Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ . = 0 by L’Hopital Rule n Let’s find the limit of the defining function x1/x.
  • 32. Sequences Example D. Is {an = n1/n = √n } a CG sequence? Find lim an if it’s a CG sequence. Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ . n∞ = 0 by L’Hopital Rule Hence {an = n1/n } is a convergent sequence and lim an = 1. n Let’s find the limit of the defining function x1/x.
  • 33. Sequences Example D. Is {an = n1/n = √n } a CG sequence? Find lim an if it’s a CG sequence. Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ . n∞ = 0 by L’Hopital Rule Hence {an = n1/n } is a convergent sequence and lim an = 1. n Let’s find the limit of the defining function x1/x. Example E. a. Show that {an = Ln(n)/n1/2} →0.
  • 34. Sequences Example D. Is {an = n1/n = √n } a CG sequence? Find lim an if it’s a CG sequence. Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ . n∞ = 0 by L’Hopital Rule Hence {an = n1/n } is a convergent sequence and lim an = 1. n Let’s find the limit of the defining function x1/x. Example E. a. Show that {an = Ln(n)/n1/2} →0. Lim Ln(x)/x1/2 lim x–1 ½ x–1/2 L‘Hospital Rule →
  • 35. Sequences Example D. Is {an = n1/n = √n } a CG sequence? Find lim an if it’s a CG sequence. Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ . n∞ = 0 by L’Hopital Rule Hence {an = n1/n } is a convergent sequence and lim an = 1. n Let’s find the limit of the defining function x1/x. Example E. a. Show that {an = Ln(n)/n1/2} →0. Lim Ln(x)/x1/2 lim = lim 2x–1/2 = 0.x–1 ½ x–1/2 L‘Hospital Rule →
  • 36. Sequences Example D. Is {an = n1/n = √n } a CG sequence? Find lim an if it’s a CG sequence. Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ . n∞ = 0 by L’Hopital Rule Hence {an = n1/n } is a convergent sequence and lim an = 1. n Let’s find the limit of the defining function x1/x. Example E. a. Show that {an = Ln(n)/n1/2} →0. Lim Ln(x)/x1/2 lim = lim 2x–1/2 = 0.x–1 ½ x–1/2 Hence {an = Ln(n)/n1/2} is a CG sequence. L‘Hospital Rule →
  • 37. Sequences Example D. Is {an = n1/n = √n } a CG sequence? Find lim an if it’s a CG sequence. Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ . n∞ = 0 by L’Hopital Rule Hence {an = n1/n } is a convergent sequence and lim an = 1. n Let’s find the limit of the defining function x1/x. Example E. a. Show that {an = Ln(n)/n1/2} →0. Lim Ln(x)/x1/2 lim = lim 2x–1/2 = 0.x–1 ½ x–1/2 Hence {an = Ln(n)/n1/2} is a CG sequence. b. Show that {bn = n20/en} is a CG sequence. L‘Hospital Rule →
  • 38. Sequences Example D. Is {an = n1/n = √n } a CG sequence? Find lim an if it’s a CG sequence. Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ . n∞ = 0 by L’Hopital Rule Hence {an = n1/n } is a convergent sequence and lim an = 1. n Let’s find the limit of the defining function x1/x. Example E. a. Show that {an = Ln(n)/n1/2} →0. Lim Ln(x)/x1/2 lim = lim 2x–1/2 = 0.x–1 ½ x–1/2 Hence {an = Ln(n)/n1/2} is a CG sequence. b. Show that {bn = n20/en} is a CG sequence. L‘Hospital Rule Applying the L‘Hospital Rule and differentiate 21 times we have [x20](21) = 0, so that {n20/en} → 0 converges. →
  • 39. Sequences As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞, but at different paces as shown in example E.
  • 40. Sequences As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞, but at different paces as shown in example E. L’Hopital Rule gives us the following general results, in short, as x→∞:
  • 41. Sequences As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞, but at different paces as shown in example E. L’Hopital Rule gives us the following general results, in short, as x→∞: * Ln(x) goes to ∞ “slower” than xp
  • 42. Sequences As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞, but at different paces as shown in example E. L’Hopital Rule gives us the following general results, in short, as x→∞: * Ln(x) goes to ∞ “slower” than xp and * xp goes to ∞ “slower” than ex.
  • 43. Sequences As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞, but at different paces as shown in example E. L’Hopital Rule gives us the following general results, in short, as x→∞: * Ln(x) goes to ∞ “slower” than xp and * xp goes to ∞ “slower” than ex. (Sin(x) and cos(x) are bounded between –1 and 1.)
  • 44. Sequences As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞, but at different paces as shown in example E. L’Hopital Rule gives us the following general results, in short, as x→∞: * Ln(x) goes to ∞ “slower” than xp and * xp goes to ∞ “slower” than ex. (Sin(x) and cos(x) are bounded between –1 and 1.) Here is a useful theorem for justifying convergence.
  • 45. Sequences As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞, but at different paces as shown in example E. L’Hopital Rule gives us the following general results, in short, as x→∞: * Ln(x) goes to ∞ “slower” than xp and * xp goes to ∞ “slower” than ex. (Sin(x) and cos(x) are bounded between –1 and 1.) n∞ n∞ (The Sandwich Theorem) Let 0 ≤ an ≤ bn for all but finitely many n’s and that lim bn = 0, then lim an = 0. Here is a useful theorem for justifying convergence.
  • 46. Sequences As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞, but at different paces as shown in example E. L’Hopital Rule gives us the following general results, in short, as x→∞: * Ln(x) goes to ∞ “slower” than xp and * xp goes to ∞ “slower” than ex. (Sin(x) and cos(x) are bounded between –1 and 1.) n∞ n∞ (The Sandwich Theorem) Let 0 ≤ an ≤ bn for all but finitely many n’s and that lim bn = 0, then lim an = 0. The Sandwich Theorem offers another method in justifying convergence besides by the L’Hopital Rule. Here is a useful theorem for justifying convergence.
  • 47. Sequences Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
  • 48. Sequences n∞ Example F. Justify that lim an = 10n/n! = 0 Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
  • 49. Sequences n∞ Example F. Justify that lim an = 10n/n! = 0 Compare bn = {1011/n} to an = {10n/n!}. Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
  • 50. Sequences n∞ Example F. Justify that lim an = 10n/n! = 0 We will show that bn > an (> 0) for sufficiently large n, Compare bn = {1011/n} to an = {10n/n!}. and since bn0 we must have an0. Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
  • 51. Sequences n∞ Example F. Justify that lim an = 10n/n! = 0 We will show that bn > an (> 0) for sufficiently large n, Compare bn = {1011/n} to an = {10n/n!}. and since bn0 we must have an0. Specifically we claim that bn > an for 10 < n = 11,12,.. Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
  • 52. Sequences n∞ Example F. Justify that lim an = 10n/n! = 0 We will show that bn > an (> 0) for sufficiently large n, For n > 10, note the following blocks in their products 10n n! = 10 * 10….. 10*10*10*10…10 n * (n–1)…11*10* 9 * 8…..1 Compare bn = {1011/n} to an = {10n/n!}. and since bn0 we must have an0. Specifically we claim that bn > an for 10 < n = 11,12,.. : Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1) an =
  • 53. Sequences n∞ Example F. Justify that lim an = 10n/n! = 0 We will show that bn > an (> 0) for sufficiently large n, For n > 10, note the following blocks in their products 10n n! = 10 * 10….. 10*10*10*10…10 n * (n–1)…11*10* 9 * 8…..1 1 Compare bn = {1011/n} to an = {10n/n!}. and since bn0 we must have an0. Specifically we claim that bn > an for 10 < n = 11,12,.. : Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1) an =
  • 54. Sequences n∞ Example F. Justify that lim an = 10n/n! = 0 We will show that bn > an (> 0) for sufficiently large n, For n > 10, note the following blocks in their products 10n n! = 10 * 10….. 10*10*10*10…10 n * (n–1)…11*10* 9 * 8…..1 10101 Compare bn = {1011/n} to an = {10n/n!}. and since bn0 we must have an0. Specifically we claim that bn > an for 10 < n = 11,12,.. : Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1) an =
  • 55. Sequences n∞ Example F. Justify that lim an = 10n/n! = 0 We will show that bn > an (> 0) for sufficiently large n, For n > 10, note the following blocks in their products 10n n! = 10 * 10….. 10*10*10*10…10 n * (n–1)…11*10* 9 * 8…..1 10101 < 1011 n Compare bn = {1011/n} to an = {10n/n!}. and since bn0 we must have an0. Specifically we claim that bn > an for 10 < n = 11,12,.. : Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1) an = = bn
  • 56. Sequences n∞ Example F. Justify that lim an = 10n/n! = 0 We will show that bn > an (> 0) for sufficiently large n, For n > 10, note the following blocks in their products 10n n! = 10 * 10….. 10*10*10*10…10 n * (n–1)…11*10* 9 * 8…..1 10101 < 1011 n n∞ Compare bn = {1011/n} to an = {10n/n!}. and since bn0 we must have an0. Specifically we claim that bn > an for 10 < n = 11,12,.. : 0 Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1) an = = bn
  • 57. Sequences n∞ Example F. Justify that lim an = 10n/n! = 0 We will show that bn > an (> 0) for sufficiently large n, For n > 10, note the following blocks in their products 10n n! = 10 * 10….. 10*10*10*10…10 n * (n–1)…11*10* 9 * 8…..1 10101 < 1011 n So for large n, bn = 1011/n > an = 10n/n!. n∞ Compare bn = {1011/n} to an = {10n/n!}. and since bn0 we must have an0. Specifically we claim that bn > an for 10 < n = 11,12,.. : 0 Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1) an = = bn
  • 58. Sequences n∞ Example F. Justify that lim an = 10n/n! = 0 We will show that bn > an (> 0) for sufficiently large n, For n > 10, note the following blocks in their products 10n n! = 10 * 10….. 10*10*10*10…10 n * (n–1)…11*10* 9 * 8…..1 10101 < 1011 n So for large n, bn = 1011/n > an = 10n/n!. Since bn0, therefore lim10n/n! = 0 by the sandwich theorem. n∞ Compare bn = {1011/n} to an = {10n/n!}. and since bn0 we must have an0. Specifically we claim that bn > an for 10 < n = 11,12,.. : 0 n∞ Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1) an = = bn
  • 60. Sequences Properties of CG Sequences Let lim an = L and lim bn = K be CG sequences, 1. If c is a constant, {can} converges and lim can = cL.
  • 61. Sequences Properties of CG Sequences Let lim an = L and lim bn = K be CG sequences, 1. If c is a constant, {can} converges and lim can = cL. 2. {an± bn}, {an bn} and {an/bn} converge
  • 62. Sequences Properties of CG Sequences Let lim an = L and lim bn = K be CG sequences, 1. If c is a constant, {can} converges and lim can = cL. 2. {an± bn}, {an bn} and {an/bn} converge with lim (an ± bn) = L ± K, lim anbn = (lim an)(lim bn) = LK, and lim (an / bn) = (lim an) / (lim bn) = L / K if K ≠ 0.
  • 63. Sequences Properties of CG Sequences Let lim an = L and lim bn = K be CG sequences, 1. If c is a constant, {can} converges and lim can = cL. 2. {an± bn}, {an bn} and {an/bn} converge with lim (an ± bn) = L ± K, lim anbn = (lim an)(lim bn) = LK, and lim (an / bn) = (lim an) / (lim bn) = L / K if K ≠ 0. Example G. Given that {an = 2 + 1/n} → 2, {bn = 3cos(1/n)} → 3 as n →∞, then
  • 64. Sequences Properties of CG Sequences Let lim an = L and lim bn = K be CG sequences, 1. If c is a constant, {can} converges and lim can = cL. 2. {an± bn}, {an bn} and {an/bn} converge with lim (an ± bn) = L ± K, lim anbn = (lim an)(lim bn) = LK, and lim (an / bn) = (lim an) / (lim bn) = L / K if K ≠ 0. Example G. Given that {an = 2 + 1/n} → 2, {bn = 3cos(1/n)} → 3 as n →∞, then * {10an = 10(2 + 1/n)} → 10(2) = 20
  • 65. Sequences Properties of CG Sequences Let lim an = L and lim bn = K be CG sequences, 1. If c is a constant, {can} converges and lim can = cL. 2. {an± bn}, {an bn} and {an/bn} converge with lim (an ± bn) = L ± K, lim anbn = (lim an)(lim bn) = LK, and lim (an / bn) = (lim an) / (lim bn) = L / K if K ≠ 0. Example G. Given that {an = 2 + 1/n} → 2, {bn = 3cos(1/n)} → 3 as n →∞, then * {10an = 10(2 + 1/n)} → 10(2) = 20 * {an * bn = (2 + 1/n) * 3cos(1/n)}
  • 66. Sequences Properties of CG Sequences Let lim an = L and lim bn = K be CG sequences, 1. If c is a constant, {can} converges and lim can = cL. 2. {an± bn}, {an bn} and {an/bn} converge with lim (an ± bn) = L ± K, lim anbn = (lim an)(lim bn) = LK, and lim (an / bn) = (lim an) / (lim bn) = L / K if K ≠ 0. Example G. Given that {an = 2 + 1/n} → 2, {bn = 3cos(1/n)} → 3 as n →∞, then * {10an = 10(2 + 1/n)} → 10(2) = 20 * {an * bn = (2 + 1/n) * 3cos(1/n)} → (2)(3) = 6
  • 67. Sequences Properties of CG Sequences Let lim an = L and lim bn = K be CG sequences, 1. If c is a constant, {can} converges and lim can = cL. 2. {an± bn}, {an bn} and {an/bn} converge with lim (an ± bn) = L ± K, lim anbn = (lim an)(lim bn) = LK, and lim (an / bn) = (lim an) / (lim bn) = L / K if K ≠ 0. Example G. Given that {an = 2 + 1/n} → 2, {bn = 3cos(1/n)} → 3 as n →∞, then * {10an = 10(2 + 1/n)} → 10(2) = 20 * {an * bn = (2 + 1/n) * 3cos(1/n)} → (2)(3) = 6 * {an / bn = (2 + 1/n) / 3cos(1/n)} → 2/3.
  • 68. Sequences A non-convergent sequence is a called a divergent (DG) sequence. Divergent Sequences
  • 69. Sequences A non-convergent sequence is a called a divergent (DG) sequence. In other words, a sequence {an} diverges (DG) if lim an is UDF.n∞ Divergent Sequences
  • 70. Sequences A non-convergent sequence is a called a divergent (DG) sequence. In other words, a sequence {an} diverges (DG) if lim an is UDF.n∞ Divergent Sequences A nice sequence {fn} might be defined by a badly behaved function f(x).
  • 71. Sequences A non-convergent sequence is a called a divergent (DG) sequence. In other words, a sequence {an} diverges (DG) if lim an is UDF.n∞ Divergent Sequences A nice sequence {fn} might be defined by a badly behaved function f(x). Example G. The sequence 0, 0, 0,.. is defined by {fn= n*sin(n)} which match the x-intercepts of y = x*sin(x).
  • 72. Sequences A non-convergent sequence is a called a divergent (DG) sequence. In other words, a sequence {an} diverges (DG) if lim an is UDF.n∞ Divergent Sequences A nice sequence {fn} might be defined by a badly behaved function f(x). Example G. (1,0) (2,0) (3,0) (4,0) y = x*sin(x) The sequence 0, 0, 0,.. is defined by {fn= n*sin(n)} which match the x-intercepts of y = x*sin(x).
  • 73. Sequences A non-convergent sequence is a called a divergent (DG) sequence. In other words, a sequence {an} diverges (DG) if lim an is UDF.n∞ Divergent Sequences A nice sequence {fn} might be defined by a badly behaved function f(x). Example G. x→∞ (1,0) (2,0) (3,0) (4,0) y = x*sin(x) The sequence 0, 0, 0,.. is defined by {fn= n*sin(n)} which match the x-intercepts of y = x*sin(x). but lim x*sin(x) is undefined. Lim fn = 0 hence {fn= n*sin(n)} converges,
  • 74. Sequences The convergence of the sequence does not imply the convergence of the function because the sequence corresponds to discrete points which contain a lot less information than the graph itself.
  • 75. Sequences The convergence of the sequence does not imply the convergence of the function because the sequence corresponds to discrete points which contain a lot less information than the graph itself. Hence unlike the case for convergent sequences, a divergent defining function f(x) is not enough to justify that the sequence {fn} diverges.
  • 76. Sequences The convergence of the sequence does not imply the convergence of the function because the sequence corresponds to discrete points which contain a lot less information than the graph itself. Hence unlike the case for convergent sequences, a divergent defining function f(x) is not enough to justify that the sequence {fn} diverges. We note the following two types of divergences, the ones that go to ∞ and the ones that jump around
  • 77. Sequences The convergence of the sequence does not imply the convergence of the function because the sequence corresponds to discrete points which contain a lot less information than the graph itself. Hence unlike the case for convergent sequences, a divergent defining function f(x) is not enough to justify that the sequence {fn} diverges. We note the following two types of divergences, the ones that go to ∞ and the ones that jump around * To-the-Infinity Divergence
  • 78. Sequences The convergence of the sequence does not imply the convergence of the function because the sequence corresponds to discrete points which contain a lot less information than the graph itself. Hence unlike the case for convergent sequences, a divergent defining function f(x) is not enough to justify that the sequence {fn} diverges. We note the following two types of divergences, the ones that go to ∞ and the ones that jump around * To-the-Infinity Divergence The function y = x → ∞ as x →∞, so the sequence 1, 2, 3.. → +∞ as n →∞.
  • 79. Sequences The convergence of the sequence does not imply the convergence of the function because the sequence corresponds to discrete points which contain a lot less information than the graph itself. Hence unlike the case for convergent sequences, a divergent defining function f(x) is not enough to justify that the sequence {fn} diverges. We note the following two types of divergences, the ones that go to ∞ and the ones that jump around * To-the-Infinity Divergence The function y = x → ∞ as x →∞, so the sequence 1, 2, 3.. → +∞ as n →∞. The function y = –x2 → –∞ as x →∞, so the sequence –1, –4, –9.. → –∞ as n →∞.
  • 80. Sequences (Divergence →±∞ ) If the defining function f(x) of fn goes to ±∞ as x →∞, then lim fn → ±∞ as n →∞ and {fn} diverges.
  • 81. Sequences (Divergence →±∞ ) If the defining function f(x) of fn goes to ±∞ as x →∞, then lim fn → ±∞ as n →∞ and {fn} diverges. Specifically, by lim fn = ∞ we mean that for any number C, we have that fn > C for all except finitely many fn’s as shown.
  • 82. Sequences (Divergence →±∞ ) If the defining function f(x) of fn goes to ±∞ as x →∞, then lim fn → ±∞ as n →∞ and {fn} diverges. Specifically, by lim fn = ∞ we mean that for any number C, we have that fn > C for all except finitely many fn’s as shown. C (lim fn = ∞) Given any number C,
  • 83. Sequences (Divergence →±∞ ) If the defining function f(x) of fn goes to ±∞ as x →∞, then lim fn → ±∞ as n →∞ and {fn} diverges. Specifically, by lim fn = ∞ we mean that for any number C, we have that fn > C for all except finitely many fn’s as shown. C Given any number C, all except finitely many fn’s, are > C. (lim fn = ∞)
  • 84. Sequences (Divergence →±∞ ) If the defining function f(x) of fn goes to ±∞ as x →∞, then lim fn → ±∞ as n →∞ and {fn} diverges. Specifically, by lim fn = ∞ we mean that for any number C, we have that fn > C for all except finitely many fn’s as shown. C Given any number C, all except finitely many fn’s, are > C. (lim fn = ∞) a few fn’s < C
  • 85. Sequences (Divergence →±∞ ) If the defining function f(x) of fn goes to ±∞ as x →∞, then lim fn → ±∞ as n →∞ and {fn} diverges. We know the behavior of the terms of a DG sequence that goes to →±∞. They blow up. Specifically, by lim fn = ∞ we mean that for any number C, we have that fn > C for all except finitely many fn’s as shown. Ca few fn’s < C Given any number C, all except finitely many fn’s, are > C. (lim fn = ∞)
  • 86. Sequences (Divergence →±∞ ) If the defining function f(x) of fn goes to ±∞ as x →∞, then lim fn → ±∞ as n →∞ and {fn} diverges. We know the behavior of the terms of a DG sequence that goes to →±∞. They blow up. Specifically, by lim fn = ∞ we mean that for any number C, we have that fn > C for all except finitely many fn’s as shown. Ca few fn’s < C Given any number C, all except finitely many fn’s, are > C. (lim fn = ∞) * Chaotic Divergence
  • 87. Sequences (Divergence →±∞ ) If the defining function f(x) of fn goes to ±∞ as x →∞, then lim fn → ±∞ as n →∞ and {fn} diverges. We know the behavior of the terms of a DG sequence that goes to →±∞. They blow up. Specifically, by lim fn = ∞ we mean that for any number C, we have that fn > C for all except finitely many fn’s as shown. Ca few fn’s < C Given any number C, all except finitely many fn’s, are > C. (lim fn = ∞) The sequence {sin(n)} = {sin(1)≈0.84, sin(2)≈0.90, sin(3)≈0.14, ..} diverges because the terms jump around between –1 and 1 without approaching a limit. * Chaotic Divergence
  • 88. Sequences Note that for any number C between –1 and 1, there are infinitely many sin(n)’s that are less than C and there are also infinitely many sin(n)’s that are more than C.
  • 89. Sequences Note that for any number C between –1 and 1, there are infinitely many sin(n)’s that are less than C and there are also infinitely many sin(n)’s that are more than C. So unlike the case where lim fn = ∞, we can’t predict the behavior of {sin(n)} as n →∞.
  • 90. Sequences Note that for any number C between –1 and 1, there are infinitely many sin(n)’s that are less than C and there are also infinitely many sin(n)’s that are more than C. So unlike the case where lim fn = ∞, we can’t predict the behavior of {sin(n)} as n →∞. Hence {sin(n)} diverges chaotically.
  • 91. Sequences Note that for any number C between –1 and 1, there are infinitely many sin(n)’s that are less than C and there are also infinitely many sin(n)’s that are more than C. So unlike the case where lim fn = ∞, we can’t predict the behavior of {sin(n)} as n →∞. Hence {sin(n)} diverges chaotically. The oscillating behavior for chaotic divergence is often caused by the periodic factors of sine or cosine which equal to 0 periodically.
  • 92. Sequences Note that for any number C between –1 and 1, there are infinitely many sin(n)’s that are less than C and there are also infinitely many sin(n)’s that are more than C. So unlike the case where lim fn = ∞, we can’t predict the behavior of {sin(n)} as n →∞. Hence {sin(n)} diverges chaotically. The oscillating behavior for chaotic divergence is often caused by the periodic factors of sine or cosine which equal to 0 periodically. Algebraic functions, Ln(x), ex, or xp (p > 0) are not periodic so sequences defined by them converge, or diverge to ±∞ (not chaotically).
  • 93. For example, {gn= n*sin(n/2)} = {1, 0, –3, 0, 5, 0, –7..} does not converge so lim x*sin(x/2) must be UDF. Sequences Summary: * a sequence {fn} inherits the “good” behaviors such as lim f(x) = L or ±∞ of its defining function f(x). * a function f(x) inherits the “bad” behavior of {fn} so if lim f(n) doesn't exist then lim f(x) doesn't exist For example, lim x/(x + 1) = 1 hence lim n/(n + 1) = 1 x→∞ n∞ * a CG sequence {fn} might be defined by a non-convergent function f(x) as x →∞. For example, the CG sequence 0, 0, 0.. is defined by {fn= n*sin(n)} but y = x*sin(x) does not converge.