Thomas Calculus 13th [Solutions]

750 Chapter 10 Infinite Sequences and Series
n
(b) Since 1 1
n
n
1 1
n
| a n|
is the sum of two absolutely convergent series,
3 2 3 2
n 1 n 1 n 1 n 1
we can rearrange the terms of the original series to find its sum: 1 1 1 1 1 1
3 9 27 2 4 8
1
1 1
3 2
1
1
1
3 2
1 1 1
2 2
60. s 1 1 1 1 1 1 1
20 1 0.6687714032 s
2 3 4 19 20 20 0.692580927
2 21
j 1 n 1 n 3
61. The unused terms are ( 1) a j ( 1) an 1 an 2 ( 1) an 3 an
4
j n 1
n 1
( 1) an 1 an 2 an 3 a n 4 . Each grouped term is positive, so the remainder has the same
n 1
sign as ( 1) , which is the sign of the first unused term.
62. sn n
n
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 2 2 3 3 4 n( n 1) k( k 1) k k 1 2 2 3 3 4 4 5 n n 1
k 1 k 1
n
which are the first 2n terms of the first series, hence the two series are the same. Yes, for s 1 1
n k k 1
k 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1
2 2 3 3 4 4 5 1 1 1 s
n n n n n n lim 1 n n
n 1
1
both series converge to 1. The sum of the first 2n 1 terms of the first series is 1 1
n 1 n 1
1.
Their sum is s lim 1 1
n
n n
n 1
1.
63. Theorem 16 states that
n
| a n|
converges
1
n
a n converges. But this is equivalent to
1
n
a n diverges
1
n
| a n|
diverges
1
64. a1 a2 an
a1 a2
a n for all n; then
n
a n converges
1
n
a n converges and these imply
1
that
n
a
n
1 n 1
a
n
65. (a)
an
b n converges by the Direct Comparison Test since an bn an b n and hence
n 1
converges absolutely
n
1
a
n
b
n
(b)
n
b n converges
1
n
1
b n converges absolutely; since
n
a n converges absolutely and
1
n
1
b
n
converges absolutely, we have
n
an ( bn ) an b n converges absolutely by part (a)
1 n 1
(c)
n
a n converges
1
k an
ka n converges
n 1 n 1
n
ka n converges absolutely
1
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