211-2008&amp;2009-3-M10-July2009

# 211-2008&amp;2009-3-M10-July2009 - KUWAIT UNIVERSITY...

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Unformatted text preview: KUWAIT UNIVERSITY Department of Mathematics 85 Computer Science Math 211 First Midterm Exam 18 July 2009 1. State whether True or False and justify your answers: (a) If Z on is convergent, then 2 cosa.ﬂ is also convergent. (b) The series 220:3[1’1—e — e'“) converges. 2. Find the limit of the sequence {on} with {11 = 2 and 3a:_lan=26ti_l —s, n >2 given that it is convergent. CO 3. Find the value oft such that 2 (%)n = 3 11:1 4. Given a sequence {on} such that :11 = 2 and 4n+101 >1 3n+2 0*” ” Gn+1 = (a) Show that Z a.“ is a divergent series. (b) Find the radius of convergence of the power series 00 2%{31' + 2)“ 11:0 5. Find the MacLaurin series for the following function: 1.5 fi3l=ma [55"{1 [3 pts] [2 pts] [3 pts] [5 pts] [3 pts] 6. Test the following series for absolute convergence, conditional convergence, or divergence: (sagas-mg (b)Z(—J“””+1 («\$2 2n+3 11:0 n20 [9 pts] COS TNT 00 CO Math 211 First Midterm Solutions Summer 2009 1. State whether True or False and justify your answers: (a) If E a." is convergent, then 2 cos on is also convergent. False Since Z: or, is convergent we have lirn on = 0 implying lirnoosc." = l at 0, hence the second series becomes divergent b} NTT. (b) The series 2:3(11—5 — e‘“) converges. True n E III—‘3 is a p—series with p = e I> 1, hence convergent; Ze‘ is a convergent geometric series since 1' = —1 4: 1. The difference of two convergent series is convergent. 2. Find the limit of the sequence {on} with ((-1 = 2 and 3c:2 an =2ai_1 —s, n 22 n—l given that it is convergent. Since the sequence is convergent, its limit L exists and we have 3L3 2 2L3 — 8 implying L = —2 . 3. Find the value oft such that f: (%)r = 3 11:1 Letr=tﬂt+2)_. thenzzozlrn=—l+2::0rn=—1+1f(1—r)=3 —r lﬂl—r): £1- —r r=3f4 —r i=6. 4. lGiven a. sequence {on} such that 0.1 = 2 and tin. + 101 35": + 2 (a) Show that Z on is a. divergent series. Gn+1= an: ”21 Applying the ratio test, we ﬁnd that lirnan+1fon = lini(4n + 101);’(3n + 2) = 4/3 > 1, which renders the series divergent. (b) Find the radius of convergence of the power series 23:0 an (3.7. + 2)" Again, ratio test gives |3\$ + 2| 4. 3/4 —P la: + 21/3I < lfil. Radius ofconv. is 1/4. 5. Find the MacLsurin series for the following function: \$5 Jail“) = W, ll'l < 1 Write the function as a derivative: 3:5 :32 115."3 1'2 d l :52 d m n 4” ﬁx) (H34)? 4 (1+x4)2 _ 4 do: 1+x4 _ 4 d3 EDP) I _’ 2 co co Jetr) = _ 3:? Z(_)r. 4” 3545—1 2 Z(_)n—1n\$4n+1 n=1 ”:1 UN 99 6. Test the following series for ebmlute convergence, conditional convergence, or divergence: m _ 111' no n ﬁn -|— 1 m e—r' (3.) “2:31:11 5111; (b) TEE—:1 2n+3 (c) g] 1+e_r_ 0051111 (5.) Divergent byNTT: . . 1'1' . sinn'fn 11111111, 5111; —'11' hm ”/71 —'11'5Eﬂ (13) CO by AST and LC-‘T: d 11‘s: + 1 1 2:: + 3 E 2I+3 _ (2x+3)212\/(I+1)_2”+11 _ 1 2:1." -I— 3 — 4(1‘ + 1:] _ (2x +3)2 2Vf(\$ +1) _ —2.1' — 1 < I] _ 2 (2m + 33121.1(;1: +1] _ 1m -|— 1 1.1111 2 0 11—1-30 23]. + 3 So the series is convergent by AST, but. it is not A0, for if we take __v'lﬁ_ 1 _en_1_ 1f11+12ﬁ_ b“‘E_ 21in _’ hmE_lm 2n+3 1 ‘1 and since 2 b" is divergent, we have 2 |c1-n| is also divergent by LOT. (0) AG by BOT: m e'” m 13—" m 1 I 1 1 e Z 1+e—n mm =2 1+e—n =Ze“+1 (Xe—‘1: 1—1/1; =e—1 r.:0 11:0 11:0 11:0 .9 CO ...
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