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Unformatted text preview: 12 CI INFINITE SEQUENCES AND SERIES 12.1 Sequences M) A sequence is an ordered list of numbers. It can also be deﬁned as a function whose domain is the set of positive integers. (b) The terms an approach 8 as n becomes large. In fact, we can make an as close to 8 as we like by taking n sulﬁciently large. (c) The terms an become large as 71 becomes large. In fact, we can make an as large as we like by taking n sufﬁciently large. 2. (a) From Deﬁnition 1, a convergent sequence is a sequence for which lim an exists. Examples: {1 / n}, {1 / 2“} Raw (b) A divergent sequence is a sequence for which lim an does not exist. Examples: {n}, {sin n} ’71—’00 1 W (0.2)", so the sequence is {0.8, 0.96, 0.992, 0.9984, 0.99968, . . }.. W _n+1 sotheseuenceis gééii —— 1&1—5—233
'a"_3n—1’ q 2’5’8’11’14"" _ ’5’2’11’7""' 3(wl) ,sotheseQuenceis{:§ § :E —3 3 } = {_3727_%7l —i } n! 9"
E 3
 15.0m =246  (2n),sothesequenceis {2,2~4,246,246c8,24c6810,...} = {2,8,48,384,3840,...}.
7. a1 = 3, an+1 : 2a,L — 1. Each term is deﬁned in terms of the preceding term.
a2=2a1—1=2(3)—1=5.a3=2a2—1=2(5)—1:9.a4=2a3—1=2(9)—1:17.
a5 : 2a4  1 = 2(17) — 1 = 33. The sequence is {3, 5,9, 17, 33, . . .}. 8. a1 = 4, an“ 2 a a" 1 Each term is deﬁned in terms of the preceding term.
4 4 4 3 4 3 .
(12 (11a: 1  :1_—1 — g. (13 — (120: 1 — g i 1  ﬁ ~ 4. Since 113 = a1, we can see that the terms ofthe sequence
. . 4 4
Will alternately equal 4 and 4/3, so the sequence IS {4, 5 , 4, E , 4, . . .}.
9. {1, %, g, %, g, . . .}. The denominator of the nth term is the nth positive odd integer, so an = 2711— 1'
1D. {1, %, g, 51;, i, . . H} The denominator ofthe nth term is the (n — 1)st power of3, so an 2 3—7113. 11. {2, 7, 12, 17, . . .}. Each term is larger than the preceding one by 5, so (111 = a1 + d(n — 1) = 2 + 5(n — 1) 2 5n  3. 12. {—%, %, —%, ﬁ, . . .} The numerator of the nth term is n and its denominator is (n + 1)2. Including the alternating signs,
71
we eta,n = ~1 "—‘——.
g ( ) (n + D2
/' \‘\ 1
13. {1, —§, %, —%, . . .}. Each term is —§ times the preceding one, so an 2 (_%)n~ . 935 i
I
I 935 :l CHAPTER12 INFINITESEQUENCESANDSERIES Z 2 r ‘ 14_ {57 1, 5, 17 5. 1, . . .}. The average of5 and 1 is 3, so we can think of the sequence as alternately adding 2 and —2 to 3
Thus, (Ln = 3 +(_1)n+1 '2 . f _ n 1 2 3 4 5 6 . 1
15 The ﬁrst 51" terms 0 a" — 2n + 1 are E, g, 7, E’ E’ 1—3 It appﬁars that the sequence 15 approaching 5. _ n I'm 1 1
7; 1 ——— Z —
7.1530 2n+1 new 2+1/n 2 9 1 1
16. {(:os(n7r/3)}n:1 — {%, 5, 1, %, 5, 1, ;, %, 1}. The sequence does not appear to have a limit. The values will cycle through the ﬁrst six numbers in the sequence—never approaching a particular number. 1 — (0.2)“, so lim an = 1 — 0 = 1 by (9). Converges 17. an : “H00
.f\‘ n3 nVn3 1 sea 1 1
' = ::————————=—————— n«a———:
“ﬂan ”3+1 (”3+1)/n3 1+1/n3: 1+0 amt—>00. CODVCI'geS
3+5n2#(3+5n2)/n2_5+3/n2 5+0_
19' 0”: n+n2 T (n+n2)/n2 _ 1+1/n’soa"_’ 1+0‘5asnﬁoo' converges H 7' \ n3 # n3/n _ n2 . . 2 .
20 an n + 1 ‘ m n/n — ‘1 + W980 an 4 0° as n —* 0° .122." = ”and .132.“ + W) = 1. Diverges 21. Because the natural exponential function is continuous at 0, Theorem 7 enables us to write lirn an : lim 61/" = elimna‘x’U/n) 2 6° 2 1. Converges
‘~\ 3n+2 323” n . . n
(:22) an Z 5” : 5” =9(%) ,so ”1320a" :9nango (g) :90=0by(9)withr= %. Converges
27m . . (27171“) / n . 27r 271' 7r
: t l bn = l ———— 2 = _ = _ ‘   7r
23_ 1f bn 1 + 8n’ hen mgr;O ”in; (1 + 8n)/n ”lingo 1/71 + 8 8 4. Slnce tan 1s continuous at Z’ by
. 2n7r . 27m 7r
Theorem 7, ”11.11.; tan<1 + 8n) : tan (7.1520 1 + 8n) 2 tan Z = 1. Converges
24. Using the last limit law for sequences and the continuity of the square root function,
, n + 1 . n + 1 . 1 + 1 / n 1 1
' = = 1 = : — : _
”lingo an ”1:1ng 9n + ”Ergo 9n + 1 71131;) 9 + l/n \/; 3. Converges
71— 71—1
(‘1) 1 n _ (:_12__ _ 1 1
25_ an = n2 + 1 .. n+ 1/n,soO g anl —— m g 5 ~+ Oasn —> oo,soan —> ObytheSqueezeTheoremand Theorem 6. Converges (—1)"n3 n3 1 26. an = ———'—" NOW lanl = m = m —> 1 as n —* 00, but the terms ofthe sequence {an} n3 + 2712 + 1
alternate in sign, so the sequence a1, a3, a5, . . . converges to —1 and the sequence 0.2, a4, a5, . . . converges to +1. This shows that the given sequence diverges since its terms don’t approach a single real number. SECTION 12.1 SEQUENCES 937 27_ a" = cos( n / 2). This sequence diverges since the terms don’t approach any particular real number as n —> 00. The terms take on values between #1 and l. 28. an : cos(2/n). As 11 —+ co, 2/n —> 0, so cos(2/n) —» cos 0 : 1 because cos is continuous. Converges 29  _____(2n _ 1)! — ——————————(2n _ 1)! — 1 —> 0 as n —+ 00 Conver es
'“" ’ (2n + 1)! ’ (2n + 1)(2n)(2n — 1)! (2n + 1)(2n) ' g
30. 271 —> 00 as n —> 00, so since lim arctanm — %, we have 11111 arctan 2n = % Converges
$>oo "—‘00
n in —71 ——2n
31. an = 9—2::  3—— : L —> Gas 71 —+ oobecause 1 + 6‘2" —> 1 and e” — e'" —> oo. Converges
611,1 6—“ en—e—n
32‘i\ _lnn___.121__# 1 H—l—ﬂlasneoo Converes
"’n2n‘1n2+1nn_%ﬁ+1 0+1_ ' g
n2 x2 H 21: H 2
: 7126—": — .Since lim —— = lim — = lim — — —,0 it follows from Theorem 3 that lim an— — 0. Converges
e" z—voo em (E—NDO 61 lace 83'3 n—>oo U
U _ ~ncosn7r — n( 1)" (207152 35 0< 1 . 2
g 5; [s1nce 0 3 cos n g 1], 211
1
36. an 2 ln(n + 1) — Inn 2 ln(n:
._ . _ sin(1/n) . . Sin(1/a;)
37. an — nsm(1/n) — 1/” . Since 11141130 1/30 that {an} converges to 1. 33. an = v" 21+3" : lim an : 8 11111 21/" = 8 . zlimwwO/W : 8 ~ 2° n—aoo n>oo (21+3n)1/n Convergent 2 x 2
. = +— l : l +—
39 y (1 1:) :> my x 11(1 33>,so 1 cos2 n
so since lim —— — —,0 m (.31.) Since [anl = n —> 00 as n —> 00, the given sequence diverges. 2n } converges to 0 by the Squeeze Theorem. ’71—’00 2n 1 . .
> 2 111(1 + a) —> In (1) = 0 as n —> 00 because ln ls continuous. Converges = lim —— [where t = 1 / 11:] = 1, it follows from Theorem 3 : (2123n)1/n : 21/7123 : 8 ' 21/71, 50 = 8 by Theorem 7, since the function f (ac) = 21 is continuous at 0. 2
. . ln(1 +2/m) H ("55> , 2
: : = = 2
$15201” $13210 1/35 x...» —1/:1:2 111.”; 1 2/w :>
. 2 E . ln 2 71 2
11m 1 + — : 11m 6 y  ,so by Theorem 3, lim 1 + E = e . Convergent
x—>oo (z: Iii—+00 naoo
”4 “\ sin 2n —1 1
40. n = . an < and lim  —0, so 3 an 3 => lim an 2 0 by the
U 1+x/E 1.3% "—“x’lilf 1+x/E 1+ﬁ new
Squeeze Theorem. Converges
2 1 2
41. an : ln(2n2 + 1) — ln(n2 + l) = ln <2; :11) = ln (%> —> 1112 as n —> oo. Convergent l
l
1, ...
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 Spring '11
 shirley

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