Thomas Calculus 13th [Solutions]

Section 10.6 Alternating Series and Conditional Convergence 751
n
66. If a ( 1) 1
n b n , then ( 1) n 1
n
n
n 1
converges, but a 1
n b n n
diverges
n 1 n 1
67. Since
n
an
converges, a n 0 and for all n greater than some N,
a n 1 and
1
2
( a ) a . Since
n
n
an
is absolutely convergent,
n 1
Test.
n
a n converges and thus
1
n
2
( a n ) converges by the Direct Comparison
1
68. For n 2,
1 1 1
n 2
n 2n
. Thus
n
1
1 1
n 2
n
diverges by comparison with the divergent harmonic series.
69. s 1 1 1
1 , s
2 2 1 ,
2 2
s 1 1 1 1 1 1 1 1 1 1 1
3 1 0.5099,
2 4 6 8 10 12 14 16 18 20 22
s 1
4 s3 0.1766,
3
s 1 1 1 1 1 1 1 1 1 1 1
5 s4 0.512,
24 26 28 30 32 34 36 38 40 42 44
s 1
6 s5 0.312,
5
s 1 1 1 1 1 1 1 1 1 1 1
7 s6 0.51106
46 48 50 52 54 56 58 60 62 64 66
70. (a) Since a n converges, say to M, for > 0 there is an integer N 1 such that
N1 1 N1 1 N1
1
an M a .
2 n an an a
2 n a
2 n 2
n 1 n 1 n 1 n N n N n N
Also,
1 1 1
a n converges to L for 0 there is an integer N 2 (which we can choose greater than or
equal to N 1 ) such that sN 2
L
2 . Therefore, a n and s
2 N 2
L
2 .
n N1
k
(b) The series a n converges absolutely, say to M. Thus, there exists N 1 such that an
M
n 1
n 1
whenever k N 1 . Now all of the terms in the sequence b n appear in a n . Sum together all of the
N1
terms in b n , in order, until you include all of the terms a n 1 , and let N n
2 be the largest index in
N2
N2
the sum b n so obtained. Then bn
M as well b n converges to M.
n 1
n 1
n 1
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