Infinite series Lecture 4 to 7.pdf

Infinite Series
Anushaya Mohapatra
Department of Mathematics
BITS PILANI K K Birla Goa Campus, Goa
November 2, 2022
Anushaya Mohapatra (Dept. of Maths) Infinite Series November 2, 2022 1 / 27

Infinite Series:
An infinite series is the sum of an infinite sequence
{an} of numbers:
a1 + a2 + a3 + · · ·

Infinite Series:
{an} of numbers:
a1 + a2 + a3 + · · ·
In this section we want to understand the meaning of
such an infinite sum and to develop methods to
calculate it.

Infinite Series:
{an} of numbers:
a1 + a2 + a3 + · · ·
In this section we want to understand the meaning of
such an infinite sum and to develop methods to
calculate it.
In order to give meaning for the infinite sum, we just
consider the sum of the first n terms
sn = a1 + a2 + · · · + an =
n
X
k=1
ak.

Infinite Series
We define the infinite sum by
∞
X
k=1
ak = a1 + a2 + a3 + · · · = lim
n→∞
sn
whenever the later limit exists.

Infinite Series
∞
X
k=1
ak = a1 + a2 + a3 + · · · = lim
n→∞
sn
For example consider the series
P∞
k=1 1/2k−1
.

Infinite Series
∞
X
k=1
ak = a1 + a2 + a3 + · · · = lim
n→∞
sn
For example consider the series
P∞
k=1 1/2k−1
.
sn =
n
X
k=1
1/2k−1
= 2 −
1
2n−1
, therefore we can say that
∞
X
k=1
1/2k−1
= lim
n→∞

2 −
1
2n−1

= 2

Infinite Series Conti.
Given a sequence of numbers {an}, an expression of
the form
a1 + a2 + a3 + · · · + an + · · · .
is called an infinite series. The number an is called
nth term of the series.

Given a sequence of numbers {an}, an expression of
the form
a1 + a2 + a3 + · · · + an + · · · .
is called an infinite series. The number an is called
nth term of the series.
The sequence {sn} defined by
sn =
n
X
k=1
ak = a1 + a2 + · · · an
is called the sequence of partial sums of the series
and sn is called nth partial sum.

Convergence and divergence of series:

If the sequence {sn} of partial sums converges to a
limit L, we say the series converges and its sum is L.
In this case we also write
∞
X
k=1
ak = a1 + a2 + a3 + · · · = L = lim
n→∞
sn.

If the sequence {sn} of partial sums converges to a
limit L, we say the series converges and its sum is L.
In this case we also write
∞
X
k=1
ak = a1 + a2 + a3 + · · · = L = lim
n→∞
sn.
If the sequence {sn} of partial sums does not
converge, we say the the series diverges.

Geometric Series: Geometric series are series of the
form (for a, r ∈ R)
a + ar + ar2
+ · · · + arn−1
+ · · · =
∞
X
n=1
arn−1

Geometric Series: Geometric series are series of the
form (for a, r ∈ R)
a + ar + ar2
+ · · · + arn−1
+ · · · =
∞
X
n=1
arn−1
Theorem 0.1.
If |r| 1 then the above geometric series converges and
a + ar + ar2
+ · · · + arn−1
+ · · · =
∞
X
n=1
arn−1
=
a
1 − r
.
if |r| ≥ 1, the series diverges.

Theorem 0.2 (The n-th term test).
If the series
∞
X
n=1
an converges, then an → 0.

Theorem 0.2 (The n-th term test).
If the series
∞
X
n=1
an converges, then an → 0.
Examples:
(a).
∞
X
n=1
n
√
n2 + 10
. (b).
∞
X
n=1
cos nπ.
(c).
∞
X
n=2
1
4n
. (d).
∞
X
n=1

1
n(n + 1)

.

Part (a): Consider the n-th term an =
P∞
n=1
n
√
n2+10
which converges to 1
not equal to 0, hence by the n-th term test the series diverges.

P∞
n=1
n
√
n2+10
Part (b): Here an = cos nπ = (−1)n which is not a convergent sequence
by the above theorem the series is divergent.

P∞
n=1
n
√
n2+10
Part (c): Given series is geometric series with common ration r = 1/4
whose modulus value is less then 1, hence the series converges.

P∞
n=1
n
√
n2+10
Part (c): Given series is geometric series with common ration r = 1/4
whose modulus value is less then 1, hence the series converges.
Part (d): an =

1
n(n+1)

= 1
n − 1
n+1, sn = 1 − 1
n+1 converges to 1 hence
the series is convergent.

Theorem 0.3 (Algebra of Series).
Let
∞
X
n=1
an = A and
∞
X
n=1
bn = B. Then
1
∞
X
n=1
(an ± bn) =
∞
X
n=1
an ±
∞
X
n=1
bn = A ± B
2
∞
X
n=1
kan = k
∞
X
n=1
an = kA.

Theorem 0.3 (Algebra of Series).
Let
∞
X
n=1
an = A and
∞
X
n=1
bn = B. Then
1
∞
X
n=1
(an ± bn) =
∞
X
n=1
an ±
∞
X
n=1
bn = A ± B
2
∞
X
n=1
kan = k
∞
X
n=1
an = kA.
Example: (a).
∞
X
n=1
2n
+ 3n
4n
; (b).
∞
X
n=1
5n
− 3n
4n
product and quotient rule
is absent....

Part (a):
∞
X
n=1
2n
+ 3n
4n
=
X
n
(2/3)n
+ (3/4)n
=
X
n
(2/3)n
+
X
n
(3/4)n
. By geometric series test
P
n(2/3)n
and
P
n(3/4)n
are convergent so their sum is
convergent by previous theorem.

Part (a):
∞
X
n=1
2n
+ 3n
4n
=
X
n
(2/3)n
+ (3/4)n
=
X
n
(2/3)n
+
X
n
(3/4)n
. By geometric series test
P
n(2/3)n
and
P
n(3/4)n
are convergent so their sum is
convergent by previous theorem.
Part (b): Since
∞
X
n=1
5n
4n
is divergent and
∞
X
n=1
3n
4n
is
convergent, the given series is divergent.

Series of non-negative terms:
∞
X
n=1
an with an ≥ 0.

∞
X
n=1
an with an ≥ 0.
Theorem 0.4.
A series
∞
X
n=1
an of non-negative terms converges if and
only if the sequence {sn} of its partial sums are bounded
from above.

∞
X
n=1
an with an ≥ 0.
Theorem 0.4.
A series
∞
X
n=1
from above.
Example:
1 Is the series
∞
X
n=1
1
n!
convergent?

∞
X
n=1
an with an ≥ 0.
Theorem 0.4.
A series
∞
X
n=1
from above.
Example:
1 Is the series
∞
X
n=1
1
n!
convergent?Ans: Convergent.
2 Is the series
P∞
n=1
1
n (harmonic series) convergent?
NO.

What can you say about the convergence of the series
P∞
n=1
1
n2 ?

What can you say about the convergence of the series
P∞
n=1
1
n2 ? It is convergent which follows from the
following theorem.
Integral Test:
Theorem 0.5.
Let {an} be a sequence of positive terms. Suppose that
an = f (n), where f (x) is a positive, continuous,
decreasing function of x for all x ≥ N (N is a positive
integer). Then the series
∞
X
n=N
an and the integral
Z ∞
N
f (x)dx both converge or diverge.

Examples:
(a). Find the values of p for which the following series
converges
∞
X
n=1
1
np
=
1
1p
+
1
2p
+
1
3p
+ · · · +
1
np
+ · · · .

Examples:
converges
∞
X
n=1
1
np
=
1
1p
+
1
2p
+
1
3p
+ · · · +
1
np
+ · · · .
Ans: The above series is convergent if and only if p 1.
Just apply integral test with the function f (x) = 1/xp
.

Examples:
converges
∞
X
n=1
1
np
=
1
1p
+
1
2p
+
1
3p
+ · · · +
1
np
+ · · · .
.
(b). Test the convergence of
∞
X
n=1
1
1 + n2
?

Examples:
converges
∞
X
n=1
1
np
=
1
1p
+
1
2p
+
1
3p
+ · · · +
1
np
+ · · · .
.
(b). Test the convergence of
∞
X
n=1
1
1 + n2
?
Ans: it is convergent. Just apply integral test with the
function f (x) = 1/(1 + x2
).

Theorem 0.6 (The Comparison Test).
Let
P
an,
P
cn and
P
dn be series with non-negative
terms. Suppose that for some integer N
dn ≤ an ≤ cn for all n N.

Let
P
an,
P
cn and
P
1 If
P
cn converges, then
P
an also converges.

Let
P
an,
P
cn and
P
1 If
P
cn converges, then
P
an also converges.
2 If
P
dn diverges, then
P
an also diverges.

Let
P
an,
P
cn and
P
1 If
P
cn converges, then
P
an also converges.
2 If
P
dn diverges, then
P
an also diverges.
Examples:
(a) Test the convergence of
∞
X
n=1
5
5n − 1
(b) Test the convergence of
∞
X
n=1
1
n!

Theorem 0.7 (Limit Comparison Test).
Suppose that an 0 and bn 0 for all n ≥ N for some
N ∈ N.
1 If limn→∞
an
bn
= c 0, then
P
an and
P
bn both
converge or diverge.
2 If limn→∞
an
bn
= 0 and
P
bn converges then
P
an
converges.
3 If limn→∞
an
bn
= ∞ and
P
bn diverges then
P
an
diverges.

Theorem 0.7 (Limit Comparison Test).
Suppose that an 0 and bn 0 for all n ≥ N for some
N ∈ N.
1 If limn→∞
an
bn
= c 0, then
P
an and
P
bn both
converge or diverge.
2 If limn→∞
an
bn
= 0 and
P
bn converges then
P
an
converges.
3 If limn→∞
an
bn
= ∞ and
P
bn diverges then
P
an
diverges.
Examples: Test the convergence of the following:
(a).
P∞
n=1
100
10n+1, (b).
P∞
n=1
1
2n+10.

Theorem 0.8 (The Ratio Test).
Let
P
an be a series with positive terms and suppose that
lim
n→∞
an+1
an
= ρ.

Theorem 0.8 (The Ratio Test).
Let
P
lim
n→∞
an+1
an
= ρ.
Then;
1 the series converges if ρ 1,
2 the series diverges if ρ 1
3 the test is inconclusive if ρ = 1.
Test the convergence of the following:
(a).
P∞
n=1
xn
n! , (b).
P∞
n=1
(2n)!
n!n! , (c).
P∞
n=1
1
n2

Theorem 0.9 (The root test).
Let
P
lim
n→∞
n
√
an = ρ.

Theorem 0.9 (The root test).
Let
P
lim
n→∞
n
√
an = ρ.
Then;
1 the series converges if ρ 1,
2 the series diverges if ρ 1
3 the test is inconclusive if ρ = 1.
Discuss the convergence of the following:
(a).
P∞
n=1
1−n
3n−n2 , (b).
P∞
n=1
3n
n10 , (c).
P∞
n=1
n
n+2
n2
.

Alternating series
Alternating series:
A series in which the terms are alternatively positive
and negative is called alternating series.

Alternating series
Alternating series:
Any series of the form:
∞
X
n=1
(−1)n
an or
∞
X
n=1
(−1)n+1
an where an ≥ 0
is an alternating series.

Alternating series
Alternating series:
Any series of the form:
∞
X
n=1
(−1)n
an or
∞
X
n=1
(−1)n+1
an where an ≥ 0
is an alternating series.
Examples:
∞
X
n=1
(−1)n
n
,
∞
X
n=1
(−4/3)n
,
∞
X
n=1
(−1)n
√
n.

Theorem 0.10 (The alternating series test (Leibniz
Test)).
The series:
∞
X
n=1
(−1)n+1
an = a1 − a2 + a3 − a4 + · · ·
converges, if all three of the following conditions are
satisfied:

Test)).
The series:
∞
X
n=1
(−1)n+1
an = a1 − a2 + a3 − a4 + · · ·
satisfied:
1 The an’s are positive.

Test)).
The series:
∞
X
n=1
(−1)n+1
an = a1 − a2 + a3 − a4 + · · ·
satisfied:
2 The positive an’s are (eventually) non-increasing:
an ≥ an+1 for all n ≥ N, for some integer N.

Test)).
The series:
∞
X
n=1
(−1)n+1
an = a1 − a2 + a3 − a4 + · · ·
satisfied:
2 The positive an’s are (eventually) non-increasing:
an ≥ an+1 for all n ≥ N, for some integer N.
3 an → 0 as n → ∞.

Examples:
1 If p 0, then the alternating p-series
∞
X
n=1
(−1)n+1
np
= 1 −
1
2p
+
1
3p
− · · ·
converges.

Examples:
1 If p 0, then the alternating p-series
∞
X
n=1
(−1)n+1
np
= 1 −
1
2p
+
1
3p
− · · ·
converges.
2 What can you say about the converges of
∞
X
n=1
(−1)n

2 + n
8n

?

Absolute convergence:
A series
P
an is said to converge absolutely (be
absolutely convergent) if the corresponding series of
absolute values,
P
|an|, converges.

A series
P
absolute values,
P
|an|, converges.
Examples:
P∞
n=1
(−1)n
np is absolutely convergent for p 1 and for
other values of p, the series is not absolutely
convergent.

A series
P
absolute values,
P
|an|, converges.
Examples:
P∞
n=1
(−1)n
np is absolutely convergent for p 1 and for
other values of p, the series is not absolutely
convergent.
Theorem 0.11.
If the series
P
an is absolutely convergent then it is
convergent.

Conditional convergence:
If the series
P
an is convergent but not absolutely
convergent then we say that the series
P
an is
conditionally convergent.

If the series
P
P
an is
P∞
n=1
(−1)n
np is conditionally convergent for 0 p ≤ 1.

If the series
P
P
an is
P∞
n=1
(−1)n
P∞
n=1
cos(n)
n5/2

If the series
P
P
an is
P∞
n=1
(−1)n
P∞
n=1
cos(n)
n5/2 is absolutely convergent and hence it is
convergent.

If the series
P
P
an is
P∞
n=1
(−1)n
P∞
n=1
cos(n)
convergent.
P∞
n=1
sin(2n−1)π/2
n3/4

If the series
P
P
an is
P∞
n=1
(−1)n
P∞
n=1
cos(n)
convergent.
P∞
n=1
sin(2n−1)π/2
n3/4 is conditionally convergent.

Examples
Discuss whether the following series absolutely
convergence or conditionally convergence.
(a).
P∞
n=1(−1)n 1−n
3n−n2 , (b).
P∞
n=1(−1)n ln(n)
n
conditionally convergent Absolutely convergent

Conclusion: Summary of Tests
1 The n-th Term Test: Unless an → 0, the series diverges.

2 Geometric series:
P
arn
converges if |r| 1; otherwise it
diverges.

2 Geometric series:
P
arn
diverges.
3 p-series:
P 1
np
converges if p 1; otherwise it diverges.

2 Geometric series:
P
arn
diverges.
3 p-series:
P 1
np
4 Series with nonnegative terms: Try the Integral Test, Ratio
Test or Root Test. Try comparing to a known series with the
Comparison Test.

2 Geometric series:
P
arn
diverges.
3 p-series:
P 1
np
Comparison Test.
5 Series with some negative terms: Does
P
|an| converge? If
yes, so does
P
an, since absolute convergence implies
convergence.

2 Geometric series:
P
arn
diverges.
3 p-series:
P 1
np
Comparison Test.
5 Series with some negative terms: Does
P
|an| converge? If
yes, so does
P
an, since absolute convergence implies
convergence.
6 Alternating series: ±
P
(−1)n
an; (an ≥ 0) converges if the
series satisfies the conditions of the Alternating Series Test.

Examples
Determine whether the following series convergent or divergent.
1
P∞
n=1(3)n+1
(4)−n+1

Examples
1
P∞
n=1(3)n+1
(4)−n+1
Ans:Convergent
2
∞
X
n=1
1 + cos n
en

Examples
1
P∞
n=1(3)n+1
(4)−n+1
Ans:Convergent
2
∞
X
n=1
1 + cos n
en
Ans: Convergent
3
∞
X
n=1
(−1)n
√
n
3n + 5

Examples
1
P∞
n=1(3)n+1
(4)−n+1
Ans:Convergent
2
∞
X
n=1
1 + cos n
en
Ans: Convergent
3
∞
X
n=1
(−1)n
√
n
3n + 5
Ans: Conditionally Convergent
4
∞
X
n=2
ln n
n

Examples
1
P∞
n=1(3)n+1
(4)−n+1
Ans:Convergent
2
∞
X
n=1
1 + cos n
en
Ans: Convergent
3
∞
X
n=1
(−1)n
√
n
3n + 5
4
∞
X
n=2
ln n
n
Ans: Divergent
5
∞
X
n=1
n sin
1
n

Examples
1
P∞
n=1(3)n+1
(4)−n+1
Ans:Convergent
2
∞
X
n=1
1 + cos n
en
Ans: Convergent
3
∞
X
n=1
(−1)n
√
n
3n + 5
4
∞
X
n=2
ln n
n
Ans: Divergent
5
∞
X
n=1
n sin
1
n
Ans: Divergent

More examples
1
P∞
n=1
n2 ln(n)+2
n3+4

More examples
1
P∞
n=1
n2 ln(n)+2
n3+4
Ans:Divergent
2
∞
X
n=1
(−1)n tan−1
n
n2 + 1

More examples
1
P∞
n=1
n2 ln(n)+2
n3+4
Ans:Divergent
2
∞
X
n=1
(−1)n tan−1
n
n2 + 1
Ans:Absolutely Convergent
3
∞
X
n=1
(−1)n ln n
n − ln(n)

Thank you

Infinite series Lecture 4 to 7.pdf

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