### Section 10.2: Series

```Section 10.2: Series
Definition: An infinite series is a sum of the terms of an infinite sequence {an }∞
n=1 .
∞
X
an = a1 + a2 + a3 + · · · + an + · · · .
n=1
The nth partial sum of the series is given by
s n = a1 + a2 + a3 + · · · + an =
n
X
aj .
j=1
That is,
s 1 = a1
s 2 = a1 + a2
s 3 = a1 + a2 + a3
..
.
These partial sums form a new sequence {sn }∞
n=1 , called the sequence of partial sums.
Definition: The series
∞
X
an is said to converge with sum S if the sequence of partial sums
n=1
{sn }∞
n=1 converges to S. Otherwise, the series diverges.
Example: List the first four partial sums of
∞
X
1n . Does the series converge or diverge?
n=1
The partial sums of this series are
s1
s2
s3
s4
=
=
=
=
..
.
1
1+1=2
1+1+1=3
1+1+1+1=4
sn = n.
The sequence of partial sums diverges:
lim sn = lim n = ∞.
n→∞
Therefore, the series diverges.
n→∞
∞
X
Example: List the first four partial sums of
(−1)n . Does the series converge or diverge?
n=1
The partial sums for this series are
s1
s2
s3
s4
−1
−1 + 1 = 0
−1 + 1 − 1 = −1
0.
=
=
=
=
In general,
sn =
−1, n odd
0, n even
Since the limit lim sn does not exist, the series diverges.
n→∞
∞
X
1
Example: Show that the series
converges.
2n
n=1
The partial sums for this series are
s1
s2
s3
sn
1
s1 = 1 −
2
1
s2 = 1 − 2
2 1
s3 = 1 − 3
2
..
.
1
sn = 1 − n .
2
1
=
2
1 1
3
=
+ =
2 4
4
1 1 1
7
=
+ + =
2 4 8
8
..
.
1 1
1
=
+ + ··· + n
2 4
2
The sequence of partial sums is convergent
lim sn = lim
n→∞
n→∞
1
1− n
2
= 1.
Thus, the series converges with sum 1.
Example: If the nth partial sum of the series
∞
X
n=1
an is sn =
n+1
, find the sum of the series
2n + 4
and a general formula for the nth term an .
The sum of the series is
n+1
1
= .
n→∞ 2n + 4
2
lim sn = lim
n→∞
The general term an is given by
an = sn − sn−1
n+1
n
=
−
2n + 4 2n + 2
n+1
n
=
−
2(n + 2) 2(n + 1)
(n + 1)(n + 1) − n(n + 2)
=
2(n + 1)(n + 2)
1
=
.
2(n + 1)(n + 2)
Note: There are certain types of series whose sum can be computed easily, provided that the
series is convergent.
Definition: A series is called a telescoping series if there is an internal cancellation in the
partial sums.
Example: Determine whether the given series converge. If so, find the sum of the series.
∞ X
1
1
(a)
cos − cos
n
n+1
n=1
The nth partial sum of this series is
1
1
1
1
1
1
1
+ cos − cos
+ cos − cos
+ · · · + cos − cos
sn =
cos 1 − cos
2
2
3
3
4
n
n+1
1
= cos 1 − cos
.
n+1
Since
lim sn = lim cos 1 − cos
n→∞
n→∞
1
n+1
= cos 1 − 1,
the series converges with sum cos 1 − 1.
(b)
∞
X
n=1
ln
n+1
n+2
Using properties of logarithms,
X
∞
∞
X
n+1
=
[ln(n + 1) − ln(n + 2)].
ln
n
+
2
n=1
n=1
The nth partial sum of this series is
sn = (ln 2 − ln 3) + (ln 3 − ln 4) + (ln 4 − ln 5) + · · · + [ln(n + 1) − ln(n + 2)]
= ln 2 − ln(n + 2)
2
= ln
.
n+2
Since lim sn = −∞, the series diverges.
n→∞
(c)
∞
X
n=1
1
n(n + 1)
Using partial fraction decomposition,
∞
X
n=1
∞
X
1
=
n(n + 1) n=1
1
1
−
n n+1
.
The nth partial sum of this series is
1
1
1 1
1 1
1 1
1
1
sn = 1 −
−
−
−
−
.
+
+
+
+···+
= 1−
2
2 3
3 4
4 5
n n+1
n+1
Since lim sn = 1, the series converges with sum 1.
n→∞
Definition: A geometric series is a series of the form
∞
X
arn = a + ar + ar2 + ar3 + · · · ,
n=0
where a is a constant and r is called the ratio of the series.
Theorem: (Geometric Series Test)
P
n
If |r| < 1, the geometric series ∞
n=0 ar , where a 6= 0, converges with sum
a
.
1−r
If |r| ≥ 1, the series diverges.
Proof: The nth partial sum of the series is
sn = a + ar + ar2 + · · · + arn−1 .
Then rsn = ar + ar2 + ar3 + · · · + arn and
rsn − sn = ar + ar2 + ar3 + · · · + arn − (a + ar + ar2 + · · · + arn−1 )
sn (r − 1) = a(rn − 1)
a(rn − 1)
sn =
.
r−1
If |r| < 1, then
lim sn =
n→∞
a
.
1−r
If |r| > 1, then limn→∞ sn = ∞.
If r = 1, then sn = an and {sn } diverges.
If r = −1, then {sn } = {a, 0, a, 0, . . .} and limn→∞ sn does not exist.
Example: Determine whether the given series converge. If so, find the sum.
(a)
n
∞
X
1 1
n=0
2
3
Since |r| =
1
< 1, the series converges with sum
3


1 1  1 3
3
=
= .


1
2
2 2
4
1−
3
n
∞
X
2
(b)
5
7
n=1
The series can be rewritten as
n X
n+1 X
n
∞
∞
∞
X
2
2
2
10 2
5
5
=
=
.
7
7
7
7
7
n=0
n=0
n=0
Since |r| =
2
< 1, the series converges with sum
7


10  1  10 7
=
= 2.

2
7
7 5
1−
7
∞
X
23n
(c)
5n+1
n=0
The series can be rewritten as
∞
X
(23 )n
n=0
Since |r| =
(d)
5 · 5n
=
n
∞
X
1 8
n=0
5
5
.
8
> 1, the series diverges.
5
1 1 1
1
− + −
+ −···
2 4 8 16
The series can be rewritten as
X
n
∞
1
1 1 1
1
1
.
1 − + − + −··· =
−
2
2 4 8
2
2
n=0
Since |r| =
1
< 1, the series converges with sum
2


1 1  1 2
1
= .
=

1
2
2 3
3
1+
2
Example: Find all values of x for which the series
∞
X
(x − 3)n converges. Find the sum of
n=1
the series for these values of x.
By the Geometric Series Test, the series converges if |x − 3| < 1. That is,
−1 < x − 3 < 1
2 < x < 4.
For these values of x the sum is
x−3
x−3
=
.
1 − (x − 3)
4−x
Note: In the following sections, we will discuss more general series and convergence tests
convergence. The following theorem provides a quick way to determine if a series diverges.
Theorem: (The
P Divergence Test)
The series
an diverges if
lim an 6= 0.
n→∞
Proof: Suppose the series
P
an converges with sum S. Then
lim sn = S,
n→∞
lim sn−1 = S.
n→∞
It follows that
lim an = lim (sn − sn−1 ) = S − S = 0.
n→∞
We have
P shown that if
series
an diverges.
P
n→∞
an converges, then limn→∞ an = 0. Thus, if limn→∞ an 6= 0, the
Note: If lim an = 0, the Divergence Test does not provide any information.
n→∞
Example: Determine whether the following series converge or diverge.
(a)
∞
X
n=0
ln
2n + 3
5n − 7
The series diverges by the Divergence Test
2n + 3
2
lim ln
= ln
6= 0.
n→∞
5n − 7
5
(b)
∞
X
tan−1 (n)
n=1
The series diverges by the Divergence Test
lim tan−1 (n) =
n→∞
π
6= 0.
2
∞
X
n
(c)
ln n
n=1
The series diverges by the Divergence Test
n
1
= lim n = ∞.
= lim
n→∞ ln n
n→∞
n→∞ 1
n
lim
n
∞ X
1
(d)
1+
n
n=1
The series diverges by the Divergence Test
n
1
= e 6= 0.
lim 1 +
n→∞
n
```