Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

SECTION 11.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOTTESTS 0 71
41. (i) Following the hint, we get that I ani < rn for n 2: N, and ~o since the geometric series 2:::'= 1
r" converges [0 < r < 1],
the series 2::=N lanl converges as well by the Comparison Test, and hence so does 2:::'= 1 lanl, so 2::_:'= 1 an is absolutely
convergent.
(ii) If lim . y'j'aJ = L > 1, then there is an integer N such that y'j'aJ >.1 for all n 2: N, so lanl > 1 for n 2: N. Thus,
n-too -
lim an "# 0, so 2:::'= 1
an diverges by the Test for Divergence:
n~oo
(iii) Consider f ~ [diverges] and f ~ [converges]. For each sum, lim y'j'aJ = 1, so the Root Test is inconclusive·.
n =l n n=l n n--+oo
43. (a) Since 2:: an is absolutely convergent, and since la;t I ~ !ani and Ia ~ I ~. I ani (because a;t' and a~ each equal
either an or 0), we conclude by the Comparison Test that both 2:: a;t and 2:: a;;- must be absolutely convergent.
Or: Use Theorem 11 .2.8.
(b) We will show by contradiction that both 2:: a;t and 2:: a ~ must diverge. For suppose that 2:: a;t converged. Then so
'
would l::(a;t'- ~an) by Theorem 11.2.8. But l::(a;t - ~an) = 2:: [ ~(an + !ani)- ~an ] = ~ 2:: I an i, which
diverges because 2: a,. is only conditionally convergent. Hence, 2: a;t can't converge. Similarly, neither can 2:: a~.
45. Suppose that 2:: an is conditionally convergent.
(a) 2:: n 2 an is divergent: Suppose 2:: n 2 a,. converges. Then lim n 2 an = 0 by Theorem 6 in Section 11.2, so there is an
• '1. ----+ 00 ;
' 1
integer N > 0 such that n > N =} n 2 lanl < 1. For n > N, we have !ani < 2 , so 2: !ani converges by
. n n>N
comparison with the convergent p-series I;: -.;.. In other words, 2:: an converges absoluiely, contradicting the
. n>N n
assumption that 2: an is conditionally convergent. This contradiction shows that 2:: n 2 an diverges.
Remark: The same argument shows that 2: n'~' a .. diverges for any p > 1.
(b) f (- lnl)n is conditionally convergent. It converges by the Alternating Series Test, but does not converge absolutely
n =2 n n .
·[by the Integral Test, since the function f ( x) = - ln 1 is c~ntinuous , positive, and decreasing on [2, oo) and
. X X
00
1
dx 1t dx [ t ] (-l}n
- 1
- = lim - 1
- = lim ln{ln x )] = oo . Setting an = -In for n ~ 2, we find that
2
x n x t ~oo 2 x n x t ~ oo 2 n n
00 00 ( -1)"
n~
2 nan = n~ ln n
2
converges by the Alternating Series Test.
It is easy to find conditionally convergent series 2:: an such·that ~na n diverges. Tw~ examples are 2::
00 ( -1)"-1
"~
1
00 (-1t-1
and
n
fo , both of which converge by the Alternating Series Test and fai l to converge absolutely because 2:: I ani is a
p-series with p ~ 1. In both cases, 2:: nan diverges by the Test for Divergence.
n=l
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