367
Worksheet 10.2.
Infinite Series
1. Compute the partial sums S2 , S4 , and S6 of
∞
!
(−1)k k −1 .
k=1
2. Calculate S3 , S4 , and S5 , and then find the sum of S =
∞
!
n=1
"
1
1
1
1
=
−
2
4n − 1
2 2n − 1 2n + 1
3. Use Theorem 2 to prove that the series cos
1
using the identity
−1
4n2
#
1
1
1
+ cos + cos + · · · diverges.
2
3
4
368
1
1
1
4. Use the formula for the sum of a geometric series to find the sum 1 + + 2 + 3 + · · ·
5 5
5
or state that the series diverges.
5. Use the formula for the sum of a geometric series to find the sum
the series diverges.
∞
!
7 · 3n
n=0
6. Use the formula for the sum of a geometric series to find the sum
that the series diverges.
11n
or state that
∞
!
8 + 2n
n=0
5n
or state
369
Solutions to Worksheet 10.2
1. Compute the partial sums S2 , S4 , and S6 of
∞
!
(−1)k k −1 .
k=1
1
1
S2 = (−1) · 1 + (−1) · 2 = −1 + = −
2
2
S4 = (−1)1 · 1−1 + (−1)2 · 2−1 + (−1)3 · 3−1 + (−1)4 · 4−1
1 1
1 1 1
7
= S2 − + = − − + = −
3 4
2 3 4
12
7
5
−1
S6 = S4 + a5 + a6 = − + (−1) · 5 + (−1)6 · 6−1
12
7
1 1
37
=− − + =−
12 5 6
60
1
−1
2
−1
2. Calculate S3 , S4 , and S5 , and then find the sum of S =
∞
!
n=1
"
1
1
1
1
=
−
2
4n − 1
2 2n − 1 2n + 1
"
N
!
1
1
1
−
SN =
2 2n − 1 2n + 1
n=1
1
S = lim SN = .
N →∞
2
#
=
1
using the identity
−1
4n2
#
1
1
3
4
5
−
. Thus S3 = , S4 = , S5 =
and
2 2(2N + 1)
7
9
11
1
1
1
+ cos + cos + · · · diverges.
2
3
4
"
#
1
1
1
The general term is an = cos
. Since lim an = lim cos
= cos lim
= cos 0 = 1,
n→∞
n→∞
n→∞ n + 1
n+1
n+1
the general term does not converge to zero. Thus, Theorem 3 implies that the series diverges.
3. Use Theorem 3 to prove that the series cos
1
1
1
4. Use the formula for the sum of a geometric series to find the sum 1 + + 2 + 3 + · · ·
5 5
5
or state that the series diverges.
In summation notation∞we have:
! " 1 #n
1
1
1
1·
1 + + 2 + 3 + ... =
5 5
5
5
n=0
1
The common ratio is r = , so |r| < 1. By the Theorem on the Sum of a Geometric
5
370
series, the series converges to the following sum:
∞ " #n
!
1
1
5
1
=
1 = 4 =
5
4
1− 5
5
n=0
5. Use the formula for the sum of a geometric series to find the sum
∞
!
7 · 3n
n=0
11n
or state that
the series diverges.
#n
∞ "
∞
!
3
3
7 · 3n !
7
The series
is a geometric series with the common ratio r = .
n =
11
11
11
n=0
n=0
3
Since |r| < 1, the series converges. We find the sum by setting a = 7, r =
in the
11
formula of Theorem 1. This gives:
#n
∞ "
!
3
7
5
7
77
7
=
=9
3 = 8 =
11
8
8
1 − 11
11
n=0
6. Use the formula for the sum of a geometric series to find the sum
∞
!
8 + 2n
or state
n
5
n=0
that the series diverges. We rewrite the series as the sum of the following two geometric
series:
" #n !
∞
∞
∞ n
∞
∞ " #n
!
!
!
8 + 2n ! 8
2
1
2
=
+
=
8·
+
(1)
n
n
n
5
5
5
5
5
n=0
n=0
n=0
n=0
n=0
1
2
The common ratios of the series are r = and r = respectively, hence by the Theorem
5
5
on the sum of a geometric series, the series converges. For the first series, a = 8 and
1
2
r = and for the second a = 1 and r = . Hence the series converges to the following
5
5
sums:
"
#
∞
n
!
8
1
8
8·
=
1 = 4 = 10
5
1− 5
5
n=0 " #
∞
n
!
1
2
2
1
5
=
=1
2 = 3 =
5
3
3
1− 5
5
n=0
Substituting these values in (1), we derive:
∞
!
2
2
8 + 2n
=
10
+
1
=
11
5n
3
3
n=0