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

SECTION 11.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS D 69
. lan+l l . [ w n+l {n+1)4 2 "+ 1 ] . (10 n + 1) 5 . . 00 10"
13. Ji_~ ~ = J2...~ (n + 2) 42n+3 . . 10" = n~ 42 . n + 2 = 8 < 1, so the senes n~l (n + 1)42 ~+1
,
is absolutely convergent by the Ratio Test. Since the. terms of this series are positive, absolute convergence is the same as
convergence.
00
(-1)" arctann\ rr/ 2 . oo rr/ 2 rr 1 . . oo (-1)" arctann
15.
2
< - 2
, so smce 2::: - 2 = -
\
2
2::: 2 converges (p = 2 > 1), the giVen senes 2:::
2
n n n = l n n =l n n=l n
converges absolutely by the Comparison Test.
= 0 and {-ln 1 } is decreasing. Now Inn < n, so
.. = 2 Inn n -oo nn n
17. f: ( - 1 )" converges by the Alte~nating Series Test s.ince lim - 1
1
diverges by the Comparison Test. Thus,
Inn n n=2.n n = 2 nn
- 1 - > .!, and since f: .! is the divergent (partial) harmonic series, f: - 1
1
~ (-1)". di' II
w -In IS con twna y convergent.
n=2 n
00 00
icos{nrr/ 3)1 1 1 . . cos{nrr/ 3)
19.
:5
1 1
and 2::: 1
converges (use the Ratto Test), so the senes 2::: converges absolutely by the
n. n. n=l n. n=l n.
1
Comparison Test.
. . n 2 + 1 . 1 + 1/n 2 1 .
21 . lim y/i(lJ = lim - 2 2 1 = hm 2 + 1 / 2 = - 2 < 1, so the senes 2::: 2 2 n-oo n-+oo n + n-oo n n =l n + 1
Root Test.
00
( n 2 + 1 ) " .
IS absolutely convergent by the
23. lim y'jaJ = lim " (1 + .!)" 2 = lim (1 + .!)" = e > 1 [by Equation 7.4.9 (or 7.4* .9) [ ET 3.6.6] ],
n--..oo n-... oo n n~ oo n
1)"
2
00 (
so the series L;: 1 + - diverges by the Root Test.
n =l n
. .
. I an+ll . I (n + 1) 10 C!1QO"+l n! I . 100 (n + 1) 100 . 100 ( 1 )100
25. hm -- - Inn · . - lim -- -- - lim -- 1 + -
n-oo a,. -n- co (n + 1)! n 100100" .-n-oo n+ 1 n - n-co n + 1 n
= 0 · 1 = 0 < 1
oo nl00100" .
so the series 2::: is absolutely convergent by the Ratio Test.
n=l n 1
27. Use the Ratio Test with the series
1 · 3 1 · 3 · 5 1 · 3 · 5 · 7 n - 11 · 3 · 5 · · · · · (2n - 1) _ co ,._11 · 3 · 5 · · · · · (2n - 1)
1 -3!+ - 5-!- - 7! + .. · +(- 1 ) {2n - 1)! +···-n~l( - 1
) · (2n - 1)!
lim lan+ll = lim ~ ( - 1)"~1·3·5 .... ·(2n-1)[2(n+1)-l]_ (2n-1)! I
n-oo a,. n-oo (2(n+1) - 1]! (- 1)"- 1 ·1·3·5·····(2n-1)
=lim ~ (-1)(2n+l)(2n- 1)! 1 = lim _.!_=0 < 1,
n -oo (2n + 1)(2n)(2n - 1)! n-oo 2n..
so the given series is ahsolutely convergent and therefore convergent.
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