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

SECTION 11.4 THE COMPARISON TESTS D 63
n + 1 n 1 ~ n + 1 d' b · ··h ~ 1
5. -- > - - = - for all n 2: 1, soL- 1verges y companson w1t L- h' h d' b · ·
1
, w 1c Jverges ecause 1t 1s a
1
nVn nVn Jn n=l n vn n=l vn
p-series with p = ! ~ 1.
00
9" 9" ( 9 )"
oo '9n
3 + 10n 10" 10 - . n=l n=l 3 + 10
7. < - = - for all n > 1. L:: (for is a convergent geometric series (lrl'= fo < 1 ), so L:: n
converges by the Comparison Test.
lnk 1 [' l k "k .>3] ~ lnkd. b · 'h~ 1 h'h d' b ·
9. - > - for all k 2: 3 smce n • > 1 .or _ , so LJ - k 1verges y com pan son w1t LJ -k, w 1c 1verges ecause 1t
k k k = 3 k=3 •
is a p-series with p =: 1 ~ 1 (the harmonic series). Thus, f lnkk diverges since a finite number of terms doesn't affect the
convergence or divergence of a series.
I
k=l
{fk ifk kl/3 1 00 • {fk . . 00 1
11. < ~ = k 312
= k 7 16
for all k 2: 1, so L:: converges by compar1son w1th L::
716 ,
Jk3 + 4k + 3 .v k3 k=l ,jk 3 + 4k + 3 k=l k
· which converges because it is a p-series with p = ~ > 1.
arctann 7r/2 fi II ~ arctann b . . h 1r ~ 1 hi h .
13. 1. 2
< ""1.2 or a n 2: 1, so LJ ~, 2 converges y comparison w1t - 2
LJ ""1.2• w c converges because 1t is a
n . n n =l n n=l n
constant times ap-series with p = _1.2 > 1.
4''+ 1 15.
4 . 4" (4)"
n _
3 2
00
(4)"
00
(4)"
> 3n"" = 4 3 for all n 2: 1. n~l 4 3 = 4 n~l 3 is a divergent geometric series (lrl = ~ > 1), so
00 4n+l
L:: - -- diverges by the Comparison Test.
n=l 3"- 2
1 1
17. Use the Lirni~ Comparison Test with an = JTi2'+T and bn = n:
1
lim an = lim n = lim ' = 1 > 0. Since the harmonic series f .!. diverges, so does
n-oo bn n-oo Jn 2 + 1 n- --- > - ( - ) = - - or use the Test for Divergence.
1 + 3" 3n + 3" 2 3" 2 3
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