Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 11.2 SERIES ¤ 461
GRAPH, MORE, FORMT (F3).) Now under E(t)= make the assignments: xt1=t, yt1=12/(-5)ˆt, xt2=t,
yt2=sum seq(yt1,t,1,t,1). (sum and seq are under LIST, OPS (F5), MORE.) Under WIND use
1,10,1,0,10,1,-3,1,1 to obtain a graph similar to the one above. Then use TRACE (F4) to see the values.
5.
n
s n
1 1.55741
2 −0.62763
3 −0.77018
4 0.38764
5 −2.99287
6 −3.28388
7 −2.41243
8 −9.21214
9 −9.66446
10 −9.01610
The series ∞ tan n diverges, since its terms do not approach 0.
n=1
7.
n
s n
1 0.29289
2 0.42265
3 0.50000
4 0.55279
5 0.59175
6 0.62204
7 0.64645
8 0.66667
9 0.68377
10 0.69849
From the graph and the table, it seems that the series converges.
k 1 1 1√1 √n − √ = −
1 1√2
√ + −
n +1 2
n=1
so ∞
n=1
1 √n −
=1−
1
√ k +1
,
1
√ = lim 1 −
n +1 k→∞
1
√
k +1
=1.
1 √
3
+ ···+
1 √k −
1
√
k +1
9. (a) lim a 2n
n = lim
n→∞ n→∞ 3n +1 = 2 3 ,sothesequence {a n} is convergent by (11.1.1).
(b) Since lim a n = 2 6=0,theseries ∞ a
n→∞ 3 n is divergent by the Test for Divergence.
n=1
11. 3+2+ 4 + 8 + ··· is a geometric series with first term a =3andcommonratior = 2 .Since|r| = 2 < 1, theseries
3 9 3 3
converges to
a
= 3 = 3 =9.
1 − r 1−2/3 1/3
13. 3 − 4+ 16 − 64 + ··· is a geometric series with ratio r = − 4 .Since|r| = 4 > 1, the series diverges.
3 9 3 3