Answers & Hints for Selected Exercises 689(b) For p = 1 the series diverges (see Exercise 8.2.16).
(c) For p < 1, the comparison test using (b) shows divergence.- Both L = 1 and R = 1.
- (a) Suppose L < 1; choose any L < r < 1. By €-criterion for upper limit
(2.9.7), ana n +i < r for all but finitely many n, so by Thm. 8.2.10, Lan converges.
(b) Suppose L. > 1. If L. is finite, then by the c:-criterion for lower limit
(2.9.8), a~!^1 > 1 for all but finitely many n , so by Thm. 8.2.10, Lan diverges.If L. = +oo, then a~!^1 > 1 for all but finitely many n, so the series diverges.
- Modify the proof given in Exercise 8.2.37.
n- By formula (5) in the proof of the integral test (8 .2.3), f 1 n ~dx ~ L t -
k=l
~- Thus, ~ ~ In· So, bn} is bounded below. To show bn} is monotone
d ecreasmg, · fi rs t s ow h th a t ln+l - In -- n+l^1 - ln(n+l)-lnn (n+l)-n. Th en app^1 Y th e
MVT to the last fraction to see that ln( n + 1) - ln n = ~ for some c E ( n , n + 1).
.". ln+1 - in= n~l - ~ < 0. Apply the monotone convergence theorem (2.5.3).
- Show ak+i ak = ~ 2k+2' so k->oo lim k (1 -ak+i) ak = k->oo lim k (1 -^2 2k+2 k+^1 ) = .!. 2. Then
use Raabe's test.
(^45) • FlfS. t S h OW a kak +1 - (2k+2)P 2 k+ 3 , SO k->oo l" lm k (l -ak+i) a k. - k->oo^1 Im. k(2k+3)P-(2k+2)P (2k+3)P -_
rm 2v-lpkP+ t erms of d egree <p-l ink = E.. By Raabe's test the series converges
k.:.oo 2PkP+ t erms of d egree ~p-1 m k 2 '
if ~ > 1 and diverges if ~ < 1.
- Suppose that Vn E N, 0 < an < 1.
(a) By continuity of lnx, an -t 0 => ln(l - an) -t 0. By continuity of ex,
the converse is true.
(b) ( =>) Suppose Lan converges. Then an -t 0 so 1 - an -t 1. By
L'Hopital's rule, lim 1 (lx- ) = - lim(l - x) = -1. By the sequential criterion
x ->O n x x->O
for continuity of ln({-x) at 0, }~ ... ~ -ln&-an) = 1. Note that 0 < an < 1 => 0 <
1 - an < 1 => ln(l - an) < 0 => -ln(l - an) > 0. So, by the limit comparison
test, L ln(l - an) converges.
( {::::) If L ln(l - an) converges, then ln(l - an) -t 0, so 1 - an -t 1, so
an -t O. Apply L'Hopital's rule and the limit comparison test as in ( =>).
(c) Let bn = (1 - a1)(l - a2) · · · (1 - an)· Then 0 < bn < 1 and lnbn =
l:ln(l -an)· By (b), l:an diverges~ {lnbn} diverges. Since {lnbn} is mono-
tone decreasing, the only way it can diverge is to -oo. Thus, Lan diverges
~ ln bn --7 -oo ~ bn --7 0.