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

SECTION 11.5 ALTERNATING SERIES D 67
~ ( -0.8t :.._ - ,;.., ( - 0.8t
LJ ' ~57- LJ '
n=1 n. n=1 n.
~ -0.8 + o.32 - o.oss3 + o.oi 706- o.oo2 731 + o.ooo 364- o.ooo 042 ~ -o.sso7
Adding b 8 to s 7 does t:~ot change the fourth decimal place of 57, so the sum_ of the series, correct to four decimal places,
is - 0.5507.
23. The series f: ( - 1 );+ 1 satisfies (i) of the Alternating Series Test because (
1
)6 < -i and (ii)' lim -i = 0, so the
n=l n · n + 1 n n -+oo n
1 1
series is convergent. Now bs = 56
= 0.000064 > 0.00005 and b6' = 66
~ 0.00002 < 0.00005, so by the Alternating Series
Estimation Theorem, n = 5. (That is, since the 6th term is less than the desired error, we need to afld the first 5 terms to get the
sum to the desired accuracy.)
25. The series f: ( - 1 )",· satisfies (i) of the Alternating Series Test because 10
. +lt 1
)' < - 0
1
1 and (ii) lim - 1 - = 0,
n =O 10" n. " n + . 1 " n n-ioo 10" n !
so the series !s convergent. Now bs = 10
! 31
~ 0.000 167 > 0.000 005 and b4 = 10
! 41
= 0.000 004 < 0.000 005, so by
the Alternating Series Estimation Theorem, n = 4 (since the series starts with n = 0, -not n = 1). (That is, since the 5th tenn
is less than the desired error, we need to add the first 4 terms to get the sum to the desired accuracy.)
1 1
27. b4 = I = -- :::; o.ooo 025, so
8. 40,320
00
(- 1)" 3 (-1)" ' 1 1 1
n~1 (2n)! ~ 53 = n~l (2n)! = -2 + 24 - . 720 ~ ~ 0 .4 59722
Adding b 4 to S3 does not change the fourth decinlal place of 5s, so by th~ Alternating Series Estimation Theorem, the sum of
. the series, correct to four decimal places, is - 0.4597.
72 -
29. b7 = 107 = 0.000 004 9, so
00 (- 1 )n- ln2 6 (-1)n- ln2
"' "' 1 4 9 16 2s 36 ·0 067
n~
1
10" :::; 56 = n-7;:
1
10" = 10 - 100 + 1000 - 10,000 + 100,000 - 1,000,000 = · 614
Adding b 7 to 56 does not change the fourth decimal place of 56, so by the Alternating Series Estimation Theorem, the sum of
the series, correct to four decimal places, is 0.0676.
I
00 ( -1)n-l 1 1 1 1 1 1 1
31. I: = 1- - + - - - + · · · + - - - + --- + · · · . The 50th partial sum of this series is an
n = 1 n 2 3 4 49 50 51 52
underestimate, since "~ 1
( - 1 ~n - l = 5so + ( 5
1
The result can be seen geometrically in Figure 1.
1 - 1
52 ) + ( ; 3 - 1
) + ; · ·, and the te~s in parentheses are all positive. '
54 1
33. Clearly bn = - - is decreasing and eventually positive and lim bn = 0 f~r any p. So the Series converges (by the
n+p
n -+oo
Alternating Series Test) for any P. for which every bn is defined, that is, n + p :/:- 0 for n 2 1, or p is not a negative integer.
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