2. Infinite Series
It’s easy to arrive at the situation of tracking and
adding “infinitely” many numbers.
3. Infinite Series
It’s easy to arrive at the situation of tracking and
adding “infinitely” many numbers.
Take a rope of length 1, cut and remove half of it,
then cut and remove half of what’s left.
4. Infinite Series
It’s easy to arrive at the situation of tracking and
adding “infinitely” many numbers.
Take a rope of length 1, cut and remove half of it,
then cut and remove half of what’s left.
1
cut away half
cut away half
1/2
5. Infinite Series
It’s easy to arrive at the situation of tracking and
adding “infinitely” many numbers.
Take a rope of length 1, cut and remove half of it,
then cut and remove half of what’s left.
1
cut away half
cut away half
1/2
1/4
“Eventually all” the rope would be gone so in total
we removed 1/2 + 1/4+ 1/8 + 1/16.. = 1. So it’s
possible to add infinitely many terms to a finite sum.
6. Infinite Series
It’s easy to arrive at the situation of tracking and
adding “infinitely” many numbers.
Take a rope of length 1, cut and remove half of it,
then cut and remove half of what’s left.
Repeating this process of cutting and removing half
of what’s left, how much rope “will” be removed?
1
cut away half
cut away half
1/2
1/4
“Eventually all” the rope would be gone so in total
we removed 1/2 + 1/4+ 1/8 + 1/16.. = 1. So it’s
possible to add infinitely many terms to a finite sum.
7. Infinite Series
It’s easy to arrive at the situation of tracking and
adding “infinitely” many numbers.
Take a rope of length 1, cut and remove half of it,
then cut and remove half of what’s left.
Repeating this process of cutting and removing half
of what’s left, how much rope “will” be removed?
1
cut away half
cut away half
cut away half
1/2
1/4
1/8
..
8. Infinite Series
It’s easy to arrive at the situation of tracking and
adding “infinitely” many numbers.
Take a rope of length 1, cut and remove half of it,
then cut and remove half of what’s left.
Repeating this process of cutting and removing half
of what’s left, how much rope “will” be removed?
1
cut away half
cut away half
cut away half
1/2
1/4
1/8
“Eventually all” the rope would be gone so in total
we removed 1/2 + 1/4+ 1/8 + 1/16.. = 1.
..
9. Infinite Series
It’s easy to arrive at the situation of tracking and
adding “infinitely” many numbers.
Take a rope of length 1, cut and remove half of it,
then cut and remove half of what’s left.
Repeating this process of cutting and removing half
of what’s left, how much rope “will” be removed?
1
cut away half
cut away half
cut away half
1/2
1/4
1/8
“Eventually all” the rope would be gone so in total
we removed 1/2 + 1/4+ 1/8 + 1/16.. = 1. Hence it’s
possible to add infinitely many terms to a finite sum.
..
10. Infinite Series
An infinite series such as 1/2 + 1/4+ 1/8 + 1/16.. (= 1)
that adds to a finite sum, is a convergent series.
11. Infinite Series
On the other hand, given c > 0, it’s obvious that
c + c + c + c .. = = ∞.Σn=1
∞
c
An infinite series such as 1/2 + 1/4+ 1/8 + 1/16.. (= 1)
that adds to a finite sum, is a convergent series.
12. Infinite Series
On the other hand, given c > 0, it’s obvious that
c + c + c + c .. = = ∞.Σn=1
∞
c
An infinite series such as 1/2 + 1/4+ 1/8 + 1/16.. (= 1)
that adds to a finite sum, is a convergent series.
An infinite series such as c + c + c + c .. = ∞
that adds to ∞, is a divergent series.
13. Infinite Series
On the other hand, given c > 0, it’s obvious that
c + c + c + c .. = = ∞.Σn=1
∞
c
An infinite series such as 1/2 + 1/4+ 1/8 + 1/16.. (= 1)
that adds to a finite sum, is a convergent series.
An infinite series such as c + c + c + c .. = ∞
that adds to ∞, is a divergent series.
Let's define the convergence and divergence of
non-negative series where 0 ≤ an for all n.
Convergent and Divergent Series
14. Infinite Series
On the other hand, given c > 0, it’s obvious that
c + c + c + c .. = = ∞.Σn=1
∞
c
An infinite series such as 1/2 + 1/4+ 1/8 + 1/16.. (= 1)
that adds to a finite sum, is a convergent series.
An infinite series such as c + c + c + c .. = ∞
that adds to ∞, is a divergent series.
Let's define the convergence and divergence of
non-negative series where 0 ≤ an for all n.
Convergent and Divergent Series
In the rope example above we can find the total
amount of rope removed after n steps:
15. Infinite Series
On the other hand, given c > 0, it’s obvious that
c + c + c + c .. = = ∞.Σn=1
∞
c
An infinite series such as 1/2 + 1/4+ 1/8 + 1/16.. (= 1)
that adds to a finite sum, is a convergent series.
An infinite series such as c + c + c + c .. = ∞
that adds to ∞, is a divergent series.
Let's define the convergence and divergence of
non-negative series where 0 ≤ an for all n.
Convergent and Divergent Series
In the rope example above we can find the total
amount of rope removed after n steps:
since after n steps, there are 2–n of the rope left,
16. Infinite Series
On the other hand, given c > 0, it’s obvious that
c + c + c + c .. = = ∞.Σn=1
∞
c
An infinite series such as 1/2 + 1/4+ 1/8 + 1/16.. (= 1)
that adds to a finite sum, is a convergent series.
An infinite series such as c + c + c + c .. = ∞
that adds to ∞, is a divergent series.
Let's define the convergence and divergence of
non-negative series where 0 ≤ an for all n.
Convergent and Divergent Series
In the rope example above we can find the total
amount of rope removed after n steps:
since after n steps, there are 2–n of the rope left,
so 1/2 + 1/4 +...+ 1/2n = 1 – 2–n = (2n– 1)/2n is removed.
17. Infinite Series
This sum of the first n terms
is called the n’th partial sum of the series.
1/2 + 1/4 + 1/8 +..+ 1/2n = sn
18. Infinite Series
This sum of the first n terms
is called the n’th partial sum of the series.
From the above the n’th partial sum of the series
1/2 + 1/4 + 1/8 + 1/16 +.... is
sn = 1/2 + 1/4 + 1/8 +..+ 1/2n = (2n– 1)/2n
1/2 + 1/4 + 1/8 +..+ 1/2n = sn
19. Infinite Series
This sum of the first n terms
is called the n’th partial sum of the series.
From the above the n’th partial sum of the series
1/2 + 1/4 + 1/8 + 1/16 +.... is
sn = 1/2 + 1/4 + 1/8 +..+ 1/2n = (2n– 1)/2n
1/2 + 1/4 + 1/8 +..+ 1/2n = sn
As n →∞, sn is the sum of more and more terms,
and Iim sn = (2n–1)/2n = 1.
20. Infinite Series
This sum of the first n terms
is called the n’th partial sum of the series.
From the above the n’th partial sum of the series
1/2 + 1/4 + 1/8 + 1/16 +.... is
sn = 1/2 + 1/4 + 1/8 +..+ 1/2n = (2n– 1)/2n
1/2 + 1/4 + 1/8 +..+ 1/2n = sn
As n →∞, sn is the sum of more and more terms,
and Iim sn = (2n–1)/2n = 1. This we define to be the
sum of the series 1/2 + 1/4 + 1/8 +.... ≡ lim sn = 1.
21. Infinite Series
This sum of the first n terms
is called the n’th partial sum of the series.
From the above the n’th partial sum of the series
1/2 + 1/4 + 1/8 + 1/16 +.... is
sn = 1/2 + 1/4 + 1/8 +..+ 1/2n = (2n– 1)/2n
Definition: Given a series a1 + a2 + a3 + .. =
let s1 = a1, s2 = a1 + a2, s3 = a1 + a2 + a3, ...
and sn = a1 + a2 + a3. . + an, be the n’th partial sum,
Σn=1
an
1/2 + 1/4 + 1/8 +..+ 1/2n = sn
As n →∞, sn is the sum of more and more terms,
and Iim sn = (2n–1)/2n = 1. This we define to be the
sum of the series 1/2 + 1/4 + 1/8 +.... ≡ lim sn = 1.
22. Infinite Series
This sum of the first n terms
is called the n’th partial sum of the series.
From the above the n’th partial sum of the series
1/2 + 1/4 + 1/8 + 1/16 +.... is
sn = 1/2 + 1/4 + 1/8 +..+ 1/2n = (2n– 1)/2n
we say the series converges → L
if the partial – sums sn converge → L.
Definition: Given a series a1 + a2 + a3 + .. =
let s1 = a1, s2 = a1 + a2, s3 = a1 + a2 + a3, ...
and sn = a1 + a2 + a3. . + an, be the n’th partial sum,
Σn=1
an
Σn=1
an
1/2 + 1/4 + 1/8 +..+ 1/2n = sn
As n →∞, sn is the sum of more and more terms,
and Iim sn = (2n–1)/2n = 1. This we define to be the
sum of the series 1/2 + 1/4 + 1/8 +.... ≡ lim sn = 1.
23. A non–negative series has 0 ≤ an for all n,
so {sn} is a non–decreasing sequence,
hence either a1 + a2 + a3 + ... = L converges
or a1 + a2 + a3 + ... = ∞.
Infinite Series
24. A non–negative series has 0 ≤ an for all n,
so {sn} is a non–decreasing sequence,
hence either a1 + a2 + a3 + ... = L converges
or a1 + a2 + a3 + ... = ∞.
Infinite Series
Therefore the main questions about a series are
* does the series converge to L or sum to ∞?
* if it converges, what is L?
25. A non–negative series has 0 ≤ an for all n,
so {sn} is a non–decreasing sequence,
hence either a1 + a2 + a3 + ... = L converges
or a1 + a2 + a3 + ... = ∞.
Infinite Series
Sum Formulas
Therefore the main questions about a series are
* does the series converge to L or sum to ∞?
* if it converges, what is L?
There are two summations:
* the geometric sums and
* the telescoping sums
that we can actually collect into simple formulas.
26. A non–negative series has 0 ≤ an for all n,
so {sn} is a non–decreasing sequence,
hence either a1 + a2 + a3 + ... = L converges
or a1 + a2 + a3 + ... = ∞.
Infinite Series
Sum Formulas
Therefore the main questions about a series are
* does the series converge to L or sum to ∞?
* if it converges, what is L?
There are two summations:
* the geometric sums and
* the telescoping sums
that we can actually collect into simple formulas.
By far the most important sum formula is the
geometric sums.
27. Infinite Series
Σk=0
ark
= a + ar + ar2 + ar3…
A geometric sequence is a sequence of the form
a=ar0, ar1, ar2, ar3… and the corresponding series is
28. Infinite Series
Σk=0
ark
= a + ar + ar2 + ar3…
A geometric sequence is a sequence of the form
a=ar0, ar1, ar2, ar3… and the corresponding series is
From the algebraic patterns:
(1 – r)(1 + r) = 1 – r2
,
29. Infinite Series
Σk=0
ark
= a + ar + ar2 + ar3…
A geometric sequence is a sequence of the form
a=ar0, ar1, ar2, ar3… and the corresponding series is
From the algebraic patterns:
(1 – r)(1 + r) = 1 – r2
,
(1 – r)(1 + r + r2) = 1 – r3
,
30. Infinite Series
Σk=0
ark
= a + ar + ar2 + ar3…
A geometric sequence is a sequence of the form
a=ar0, ar1, ar2, ar3… and the corresponding series is
From the algebraic patterns:
(1 – r)(1 + r) = 1 – r2
,
(1 – r)(1 + r + r2) = 1 – r3
,
(1 – r)(1 + r + r2 + r3) = 1 – r4
, etc..
we obtain the general formula
(1 – r)(1 + r + r2 … + rn – 1) = 1 – rn
,
31. Infinite Series
1 + r + r2 … + rn – 1 = 1 – rn
1 – r
Σk=0
ark
= a + ar + ar2 + ar3…
A geometric sequence is a sequence of the form
a=ar0, ar1, ar2, ar3… and the corresponding series is
From the algebraic patterns:
(1 – r)(1 + r) = 1 – r2
,
(1 – r)(1 + r + r2) = 1 – r3
,
(1 – r)(1 + r + r2 + r3) = 1 – r4
, etc..
we obtain the general formula
(1 – r)(1 + r + r2 … + rn – 1) = 1 – rn
, hence
Σ
n
r k–1
= .
32. Infinite Series
1 + r + r2 … + rn – 1 = 1 – rn
1 – r
Σk=0
ark
= a + ar + ar2 + ar3…
A geometric sequence is a sequence of the form
a=ar0, ar1, ar2, ar3… and the corresponding series is
From the algebraic patterns:
(1 – r)(1 + r) = 1 – r2
,
(1 – r)(1 + r + r2) = 1 – r3
,
(1 – r)(1 + r + r2 + r3) = 1 – r4
, etc..
we obtain the general formula
(1 – r)(1 + r + r2 … + rn – 1) = 1 – rn
, hence
Σ
n
r k–1
=
In particular, if | r | < 1 so that rn → 0,
we obtain the following formulas.
.
33. Infinite Series
Σ
∞
rn
= 1 + r + r2 + …n=0
1
1 – r=
Sums of Geometric Series: For | r | < 1,
Σ
∞
arn = a(1 + r + r2 + …)
n=0
a
1 – r=
I.
II.
This formula is valid
for negative r also.
34. Infinite Series
Σ
∞
rn
= 1 + r + r2 + …n=0
1
1 – r=
Sums of Geometric Series: For | r | < 1,
Σ
∞
arn = a(1 + r + r2 + …)
n=0
a
1 – r=
I.
II.
This formula is valid
for negative r also.
Example A. Find the sum Σn=1
5
3n
35. Infinite Series
Σ
∞
rn
= 1 + r + r2 + …n=0
1
1 – r=
Sums of Geometric Series: For | r | < 1,
Σ
∞
arn = a(1 + r + r2 + …)
n=0
a
1 – r=
I.
II.
This formula is valid
for negative r also.
Example A. Find the sum Σn=1
In the expanded form, we have
5
3n
Σn=1
5
3n = 5
3
+ 5
32 + 5
33 + ..
36. Infinite Series
Σ
∞
rn
= 1 + r + r2 + …n=0
1
1 – r=
Sums of Geometric Series: For | r | < 1,
Σ
∞
arn = a(1 + r + r2 + …)
n=0
a
1 – r=
I.
II.
This formula is valid
for negative r also.
Example A. Find the sum Σn=1
In the expanded form, we have
5
3n
Σn=1
5
3n = 5
3
+ 5
32 + 5
33 + ..
5
3
(1 + 1
3 +
1
32 + … )=
Factor out the “a” to identify r,
then use the formula.
37. Infinite Series
Σ
∞
rn
= 1 + r + r2 + …n=0
1
1 – r=
Sums of Geometric Series: For | r | < 1,
Σ
∞
arn = a(1 + r + r2 + …)
n=0
a
1 – r=
I.
II.
This formula is valid
for negative r also.
Example A. Find the sum Σn=1
In the expanded form, we have
1
1 – 1/3
5
3n
Σn=1
5
3n = 5
3
+ 5
32 + 5
33 + ..
5
3
(1 + 1
3 +
1
32 + … )
= 5
3
=
Factor out the “a” to identify r,
then use the formula.
38. Infinite Series
Σ
∞
rn
= 1 + r + r2 + …n=0
1
1 – r=
Sums of Geometric Series: For | r | < 1,
Σ
∞
arn = a(1 + r + r2 + …)
n=0
a
1 – r=
I.
II.
This formula is valid
for negative r also.
Example A. Find the sum Σn=1
In the expanded form, we have
1
1 – 1/3
5
3n
Σn=1
5
3n = 5
3
+ 5
32 + 5
33 + ..
5
3
(1 + 1
3 +
1
32 + … )
= 5
3
3
2
= 5
3
= 5
2
=
Factor out the “a” to identify r,
then use the formula.
39. Infinite Series
For geometric sums starting at a different initial value,
we may shift the index and change the problem
to one that starts at 0.
40. Infinite Series
For geometric sums starting at a different initial value,
we may shift the index and change the problem
to one that starts at 0.
Example B. Find the sum Σn=4 3n–1* 7
(–2)n+2* 5
41. Infinite Series
For geometric sums starting at a different initial value,
we may shift the index and change the problem
to one that starts at 0.
Example B. Find the sum
The summation starts with n = 4 so we need to
change the summation to a new variable k,
starting from k = 0 as required in the formula.
Σn=4 3n–1* 7
(–2)n+2* 5
42. Infinite Series
For geometric sums starting at a different initial value,
we may shift the index and change the problem
to one that starts at 0.
Example B. Find the sum
The summation starts with n = 4 so we need to
change the summation to a new variable k,
starting from k = 0 as required in the formula.
Since n starts 4 so (n – 4) starts at 0.
Σn=4 3n–1* 7
(–2)n+2* 5
43. Infinite Series
For geometric sums starting at a different initial value,
we may shift the index and change the problem
to one that starts at 0.
Example B. Find the sum
The summation starts with n = 4 so we need to
change the summation to a new variable k,
starting from k = 0 as required in the formula.
Since n starts 4 so (n – 4) starts at 0.
To shift the index:
I. express the exponents in terms of (n – 4) with the
necessary adjustments.
Σn=4 3n–1* 7
(–2)n+2* 5
44. Infinite Series
For geometric sums starting at a different initial value,
we may shift the index and change the problem
to one that starts at 0.
Example B. Find the sum
Hence,
(–2)n+2 = (–2)(n–4) + 6
3n–1 = 3 (n–4) +3
The summation starts with n = 4 so we need to
change the summation to a new variable k,
starting from k = 0 as required in the formula.
Since n starts 4 so (n – 4) starts at 0.
To shift the index:
I. express the exponents in terms of (n – 4) with the
necessary adjustments.
Σn=4 3n–1* 7
(–2)n+2* 5
45. Infinite Series
Σn = 4
∞
3n–1* 7
II. rewrite the summation and set (n – 4) to k
with the summation starting at k = 0.
(–2)n+2* 5
46. Infinite Series
Σn = 4
∞
3n–1* 7
II. rewrite the summation and set (n – 4) to k
with the summation starting at k = 0.
(–2)n+2* 5
=
3 (n–4) + 3 * 7
(–2)(n–4) + 6 * 5Σn = 4
set (n – 4) as k with the
sum starting at k = 0
47. Infinite Series
Σn = 4
∞
3n–1* 7
II. rewrite the summation and set (n – 4) to k
with the summation starting at k = 0.
(–2)n+2* 5
=
3 (n–4) + 3 * 7
(–2)(n–4) + 6 * 5
=
Σn = 4
set (n – 4) as k with the
sum starting at k = 0
3 k + 3 * 7
(–2) k + 6 * 5
Σk = 0
factor out the constant multiple
48. Infinite Series
Σn = 4
∞
3n–1* 7
II. rewrite the summation and set (n – 4) to k
with the summation starting at k = 0.
(–2)n+2* 5
=
3 (n–4) + 3 * 7
(–2)(n–4) + 6 * 5
=
Σn = 4
=
set (n – 4) as k with the
sum starting at k = 0
3 k + 3 * 7
(–2) k + 6 * 5
Σk = 0
3
–2
Σk = 03 3 * 7
(–2)6 * 5
)
k
(
factor out the constant multiple
49. Infinite Series
Σn = 4
∞
3n–1* 7
II. rewrite the summation and set (n – 4) to k
with the summation starting at k = 0.
(–2)n+2* 5
=
3 (n–4) + 3 * 7
(–2)(n–4) + 6 * 5
=
Σn = 4
=
set (n – 4) as k with the
sum starting at k = 0
3 k + 3 * 7
(–2) k + 6 * 5
Σk = 0
3
–2
Σk = 03 3 * 7
(–2)6 * 5
)
k
( 1 + (–2/3) + (–2/3)2 + ..
factor out the constant multiple
50. Infinite Series
Σn = 4
∞
3n–1* 7
II. rewrite the summation and set (n – 4) to k
with the summation starting at k = 0.
(–2)n+2* 5
=
3 (n–4) + 3 * 7
(–2)(n–4) + 6 * 5
=
Σn = 4
=
set (n – 4) as k with the
sum starting at k = 0
3 k + 3 * 7
(–2) k + 6 * 5
Σk = 0
3
–2
Σk = 03 3 * 7
(–2)6 * 5
)
k
(
=
3 3 * 7
(–2)6 * 5 1
1 – (– 2/3)
1 + (–2/3) + (–2/3)2 + ..
factor out the constant multiple
=
32 * 7
26
51. Infinite Series
The seriesΣn=1
∞
r
n + p
II. The Telescoping Series:
– r
n + q
][
constants is called a telescoping series.
where p, q, and r are
52. Infinite Series
The seriesΣn=1
∞
r
n + p
II. The Telescoping Series:
– r
n + q
][
constants is called a telescoping series. They are so
named due to the cancelation of the terms in the sum.
where p, q, and r are
53. Infinite Series
The seriesΣn=1
∞
r
n + p
II. The Telescoping Series:
– r
n + q
][
constants is called a telescoping series. They are so
named due to the cancelation of the terms in the sum.
where p, q, and r are
Example C. Find Σn=1
1
n + 2
– 1
n + 4
][
54. Infinite Series
The seriesΣn=1
∞
r
n + p
II. The Telescoping Series:
– r
n + q
][
constants is called a telescoping series. They are so
named due to the cancelation of the terms in the sum.
where p, q, and r are
Example C. Find Σn=1
1
n + 2
– 1
n + 4
][
Σ
n=1
1
n + 2
– 1
n + 4 ][
55. Infinite Series
The seriesΣn=1
∞
r
n + p
II. The Telescoping Series:
– r
n + q
][
constants is called a telescoping series. They are so
named due to the cancelation of the terms in the sum.
where p, q, and r are
Example C. Find Σn=1
1
n + 2
– 1
n + 4
][
Σ
n=1
1
n + 2
– 1
n + 4 ][
= 1
3
+( – 1
5
)
56. Infinite Series
The seriesΣn=1
∞
r
n + p
II. The Telescoping Series:
– r
n + q
][
constants is called a telescoping series. They are so
named due to the cancelation of the terms in the sum.
where p, q, and r are
Example C. Find Σn=1
1
n + 2
– 1
n + 4
][
Σ
n=1
1
n + 2
– 1
n + 4 ][
= 1
3
+( – 1
5
) 1
4
( – 1
6
)
57. Infinite Series
The seriesΣn=1
∞
r
n + p
II. The Telescoping Series:
– r
n + q
][
constants is called a telescoping series. They are so
named due to the cancelation of the terms in the sum.
where p, q, and r are
Example C. Find Σn=1
1
n + 2
– 1
n + 4
][
Σ
n=1
1
n + 2
– 1
n + 4 ][
= 1
3
+( – 1
5
) 1
4
( – 1
6
)+ 1
5
( – 1
7
)+ 1
6
( – 1
8
)+ 1
7
( – 1
9
) …
58. Infinite Series
The seriesΣn=1
∞
r
n + p
II. The Telescoping Series:
– r
n + q
][
constants is called a telescoping series. They are so
named due to the cancelation of the terms in the sum.
where p, q, and r are
Example C. Find Σn=1
1
n + 2
– 1
n + 4
][
Σ
n=1
1
n + 2
– 1
n + 4 ][
= 1
3
+( – 1
5
) 1
4
( – 1
6
)+ 1
5
( – 1
7
)+ 1
6
( – 1
8
)+ 1
7
( – 1
9
) …
= 1
3
+ 1
4
= 7
12.
59. Infinite Series
The seriesΣn=1
∞
r
n + p
II. The Telescoping Series:
– r
n + q
][
constants is called a telescoping series. They are so
named due to the cancelation of the terms in the sum.
where p, q, and r are
Example C. Find Σn=1
1
n + 2
– 1
n + 4
][
Σ
n=1
1
n + 2
– 1
n + 4 ][
= 1
3
+( – 1
5
) 1
4
( – 1
6
)+ 1
5
( – 1
7
)+ 1
6
( – 1
8
)+ 1
7
( – 1
9
) …
= 1
3
+ 1
4
= 7
12.
Note that If n2 + bn + c has integer roots p and q,
1
n2+bn+c = r
n – p – r
n – q for some r.then
60. Infinite Series
The seriesΣn=1
∞
r
n + p
II. The Telescoping Series:
– r
n + q
][
constants is called a telescoping series. They are so
named due to the cancelation of the terms in the sum.
where p, q, and r are
Example C. Find Σn=1
1
n + 2
– 1
n + 4
][
Σ
n=1
1
n + 2
– 1
n + 4 ][
= 1
3
+( – 1
5
) 1
4
( – 1
6
)+ 1
5
( – 1
7
)+ 1
6
( – 1
8
)+ 1
7
( – 1
9
) …
= 1
3
+ 1
4
= 7
12.
Note that if n2 + bn + c has integer roots p and q,
1
n2+bn+c = r
n – p – r
n – q for some r.then
62. Infinite Series
Example D. Find Σn=4
1
r
n – 3 – r
n + 1
n2 – 2n – 3
n2 – 2n – 3
1 =
Since n2 – 2n – 3 = (n – 3)(n + 1), we can find r
so that
63. Infinite Series
Example D. Find Σn=4
1
Multiplying both sides by (n – 3)(n + 1),
we have 1 = r(n + 1) – r(n – 3).
r
n – 3 – r
n + 1
n2 – 2n – 3
n2 – 2n – 3
1 =
Since n2 – 2n – 3 = (n – 3)(n + 1), we can find r
so that
64. Infinite Series
Example D. Find Σn=4
1
Multiplying both sides by (n – 3)(n + 1),
we have 1 = r(n + 1) – r(n – 3).
Setting n = –1 we get 1 = 4r, or r = 1/4.
r
n – 3 – r
n + 1
n2 – 2n – 3
n2 – 2n – 3
1 =
Since n2 – 2n – 3 = (n – 3)(n + 1), we can find r
so that
65. Infinite Series
Example D. Find Σn=4
1
Multiplying both sides by (n – 3)(n + 1),
we have 1 = r(n + 1) – r(n – 3).
Setting n = –1 we get 1 = 4r, or r = 1/4.
r
n – 3 – r
n + 1
n2 – 2n – 3
n2 – 2n – 3
1 =
Since n2 – 2n – 3 = (n – 3)(n + 1), we can find r
so that
Hence we can write the series in a telescoping form:
Σn=4
1
n2 – 2n – 3 = Σ
n=4
¼
n – 3 – ¼
n + 1
66. Infinite Series
Example D. Find Σn=4
1
Multiplying both sides by (n – 3)(n + 1),
we have 1 = r(n + 1) – r(n – 3).
Setting n = –1 we get 1 = 4r, or r = 1/4.
r
n – 3 – r
n + 1
n2 – 2n – 3
n2 – 2n – 3
1 =
Since n2 – 2n – 3 = (n – 3)(n + 1), we can find r
so that
Hence we can write the series in a telescoping form:
Σn=4
1
n2 – 2n – 3 = Σ
n=4
¼
n – 3 – ¼
n + 1
= ¼(1 + 1/2 + 1/3 + 1/4)
67. Infinite Series
Example D. Find Σn=4
1
Multiplying both sides by (n – 3)(n + 1),
we have 1 = r(n + 1) – r(n – 3).
Setting n = –1 we get 1 = 4r, or r = 1/4.
r
n – 3 – r
n + 1
n2 – 2n – 3
n2 – 2n – 3
1 =
Since n2 – 2n – 3 = (n – 3)(n + 1), we can find r
so that
Hence we can write the series in a telescoping form:
Σn=4
1
n2 – 2n – 3 = Σ
n=4
¼
n – 3 – ¼
n + 1
= ¼(1 + 1/2 + 1/3 + 1/4)
= 25/48
