Answers & Hints for Selected Exercises 691(^16). Th esenes. 1 + 2 1 1 - 3 + 4 1 1 + 5 - 61 + 7 1 1 1 + 8 - 9 +··· d' ivergessmce1tssequence · · {S n }
of partial sums has the subsequence S3n = £'.: [ 3 k1 2 + 3 k1 1 - 3 \ J. S3n ---+ oo
k=l
. 1 1 1 9k2-2 8k2 8 A 1 Th 8 1 10
smce 3k-2 + 3k-1 - 3k = 3k(3k-1)(3k-2) > (3k)(9k2) = 27k · PP Y m. · · ·
18. Using a computer to check the inequalities, the first 24 t erms are 1 - ~
11111 1 1 111 1 1 111 1 1 1
- 4 - 6 - 8 + 3 - 10 - 12 - 14 - 16 + 5 - 18 - 20 - 22 - 24 + 7 - 26 - 28 - 30 - 32
1 1 1 1
- 9 - 34 - 36 - 38.
EXERCISE SET 8.4- L Ck = 1 + 0 + 0 + 0 + 0 + ... = l.
00 - For lrl < 1, the series :Z::: rk converges a bsolutely, so the Cauchy prod-
k=O
uct L Ck of this series with itself must converge to ( f rk)
2. Now ck =
k=O
itarirk-i = (k + l)rk. Thus, k~
0
(k + l)rk = C~
0rk)2
= (l~r)2. So,
00 00 00
'\""" L,, k r k + L,, '\""" r k - (l-r) (^1) 2 , fr om w h' lC h we ge t '\""" L,, k r k - (l- (^1) rJ2 - l-r (^1) - _ (l-r)2. r
k=O k=O k=O
00 00
- Let lrl < l. We know that L rk = l~r and from Ex. 3, (l!r)2 = L (k +
k=O k=O
- rk. Thus, (l!r) 3 = ( f rk) ( f (k + 1) rk). Since both of these series con-
k=O k=O
verge absolutely, their Cauchy product series :Z::: Ck will have the same sum.
k.. •. k+l. (k+l}(k+2}.
Now, Ck = :Z::: r' [(k + 1 - i) rk-•] = rk L i = r k i. Thus, (l!r)3 =
i=O i=l
~ (k+l)(k+2} k
L,, 2 r.
k=O
00 ( )k- The series L -~ converges to ln 2 (Exercise 8.3. 7), and the geometric
k=l
00 ~
series :Z::: ~ converges absolutely to 1 _ 113 = ~· Thus, by Mertens' theorem,
k=l 00
their Cauchy product series L Ck converges to the product of their sums, ~ ln 2.
k=l
Note that ck -- ~ u (-l}i - i -^1
i=l^3 k+1- i ·
- (-l}k+l - (-l)k+l
- Let ak - ~and bk-~· Then
(a) L ak converges absolutely, since L lak I = L ~, a convergent p-series.