211-2007&2008-3-M10-July2008

211-2007&2008-3-M10-July2008 - Kuwait University...

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Unformatted text preview: Kuwait University Faculty of Science Dept. of Math. 35 Comp. Sci. First Mid-Term Exam in CALCULUS C (Math7211) Time: 90 min. July 5, 2008 Calculators and mobile phones are NOT allowed. Answer the following questions: 1. [(2+2) pts.] Find the limits of the following sequences: 2. [(2+2+3) pie] Determine whether the following series is convergent or divergent, if it. is con- vergent, find its sum. (a) (b) 1T " 9 '" (c) mpg—W» 11:1 3 “2 11:1 I 11:1 n + 4n + 3 3. [3 pts.] Determine whether the follmving series is absolutely convergent, conditionally conver- gent or divergent: .4. . Z 5111 n n_1 7" + n7 4. [3 pts.] For what values of p the following series is convergent. 00 n 2 n n! em". 71:} 5. [4 pts.] Find the radius and interval of convergence of the series: 00 n (2.1: + 3 " 2H) (—i . nlnn 71:2 6. [4 pts.] (3.) Find the power series representation of the function fix) : ln(1 + 3):: m (b) Use part(e) to find the sum of the series Z: (772:3) 11:1 00 99 1a. 1b. 2a. 2b. 1 - sin l . liln (—O—Ojjfl 11111 E") = 0.Heuce, the sequence ls convergent because. 11—100 71, 114100 _ Tl 1 —— cos 251112 E . . lim (—mfl : lim 2 : O by applymg the sandwmh theorem and 113—100 71 71:00 TL ' - 1 3111 “- lim 1( n) x 1 by applying the rule. '11—‘00 '— '11 +1 2 ” 1 2 T“? M 71" n 11 11 n . _ Inn (—1)”( ) tan—1n = iim ((1 + —) ) [1111 ten 111 : 2 x 7 : 11. 71—100 71, 71—100 :17, nu-roo 2, The sequence is divergent (theorem). 1 a”: 3 2 71—1 n rule twice). 6 9/; 11m TLHDO n2 2 00 using L’Hopitai’s is divergent by applying the nth term test.( 00 Do in Z «Hm 91’” : 97r Z is convergent because it is a geometric series with Iri = 11:1 2 7121 Its sum is 5;. 20 . 2 i 2 = 1 _ 1 ' 1124-4114 3 (n+1)(n l .5) (n+1) (n+3) 1 1 1 1. 1 1 1 1 1 1 1 1 1 1 Sn : (s — r)+(s — EH 1 * s)+(s ‘ 1l+ +(11]_ m)+(s ’ whim — :47) 6a. 6b. l" . a , . . . 11111 311 = —. [‘he serles 1s convergent and 1ts sum equal % n»—ooo 6 OD . smn sin“ 1 . . . E 7“ + “7 IS A.C., for W S g = b", but 20,. IS conv. (p-series, p = 7 > 1), usmg 11:] the B.C.T., we deduce that the series is A.C. m . (11+ 1)n+1 n! e”? 1 , — (11 +1)! em“)? X 11.” (31’ HIE-go“ + vergent when 61” < 1. i.e. 1—— p < 0. Hence, 1? > 1. t . 3 n+1 1 (291+ ) 111111 : ‘2$+3‘ 11m 11 11111 (n + 1) 111(11 + 1) (2m ~+~ 3)” nfloo 11+ 1111(11.+ 1) It is conv. when |2w + 3) < 1. which implies 72 < '1; < 71. Atm: ~1; EH)” 1 11 ln 11 . a 1 11111 71+ 11—400 an 1 _ . . —)" : e‘1 P. The series 1s con- 11. lim : i2$+3|. 714100 it is conv. (by A.S.T.) n=2 °° 1 At 1' : i2 : 2 it is div. (by integral test). Thus the interval of conv. is {w}, *2) _ 1111111 n——2 and the radius is 0° 71 11—1 n+2 332111(1+ :12) : Z "(m—J—m 11 71:1 . . . g , :82 m (*1)“'1 (11. 71- 2)m"‘"H 1 lef. both Sides, we obtain 231111[1+ + 1 + I :2 n; Set :5 z :5, °° 2 1 we obtain 2 = E 1112. 11:1 00 99 ...
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This note was uploaded on 02/23/2010 for the course CAL C 0410211 taught by Professor Deparpment during the Spring '10 term at Kuwait University.

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211-2007&amp;2008-3-M10-July2008 - Kuwait University...

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