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

CHAPTER 11 REVIEW 0 101 l
23. Consider the series of absolute values: E n -I/ 3 is a p-series with p = t ~ 1 and is therefore divergent. But if we apply the
1'1=1
Alternating Series Test, we see that bn = 3~ > 0, {bn} is decreasing, and lim bn = 0, so the series E (- 1)"- 1 n - 1 / 3
yn n-Hxl n=l
converges. Thus, E (-1)"- 1 n- 1 1 3 is conditionally convergent.
tt= l
an+ll ~ ( - 1t+ 1 (n+2) 3n+l . 2 2 "+1 l - n+2 3 _ 1+(2/n) 3 3 .
/ n) · 4 --+ 4 < 1. asn--+ oo, so by the Ratro
l
25. ~ ·~ 22n+3 · (- 1 )"(n + 1 ) 3 " .- n+ 1 · 4 - 1 + ( 1
oo ( -1)"(n + 1)3"
Test, I:
2 "+ 1 is absolutely convergent.
n=l 2
oo (- 3)n- 1 oo (- 3)n-l oo (- 3)"-1 1 oo (- 3)n- l 1 oo ( 3 ) n-l 1 ( · 1 )
27· I: 23n = I: (23)n = L 8" = -8 I: 8"- 1 = -8 L - -8 = -8 1 ( 3/8)
n=I n=l n=l n=l n=l - -
1 8 1
=s· u = u
00
29. :L[tan- 1 (n+ 1)-tan- 1 n) = lim Bn
n = l
n-oo
31. 1 - e + - 21
. -
= lim [(tan- 1 2- tan- 1 1) + (tan- 1 3- tan- 1 2) + · · · + (tan- 1 (n + 1)- tan- 1 n)J
n-oo
= lim [tan- 1 (n + 1)- tan- 1 1) = 1!. - 1!. = 1!.
n-oo 2 4 4
e2 e3 e4 oo e" oo ( -e)n oo x"
31 . + - 41 . - .. · = L (-1)"1 = E -- 1- = e- e since e"' = I: I for all x.
1 · 1 ( oo x" oo (-x)" )
33. cosh x = -(e"' +e-"') = - I: 1 + L -- 1
-
2 2 n=O n. n=O n.
n=O n. n=O n. . n=O n .
oo ( -1) n+l · 1 1 1 1 1 1 1
' 35· n~l n5 = 1 - 32 + 243 - 1024 + 3125 - 7776 + 16,807- 32,768 + ... ·
. 1 1 oo (-1)"+1 7 (- 1)"+1
~ E 5 ~ 0.9721.
' n=l n n=l n
Srnce ba = 85
= 32 768
< 0.000031, I:
5
00
1 8 1 1 1
37. I: - 2 5
" ~ E - 2 5
" ~ 0.18976224. To estimate the error, note that - 2
- ·- < -,so the remainder term is
n=l + n=l + + 5" 5n .
00
1
00
1 1/5 9
Ra = n~o
2 + 5 " < n~o S" = 1 _ 115 = 6.4 ~ 10,- 7 [geometric series with a= -b and r = i ].
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