Math31B_VanKoten_Fall13_Midterm 2 solutions.pdf

Solution we compute z 1 1 x 3 2 dx lim r z r 1 1 x 3

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or show that the integral diverges. Solution : We compute Z 1 1 ( x + 3) 2 dx = lim R →∞ Z R 1 1 ( x + 3) 2 dx = lim R →∞ - 1 x + 3 x = R x =1 = lim R →∞ 1 4 - 1 R + 3 = 1 4 . (b) (10 points) Let a n = n ! n n . Either compute lim n →∞ a n or show that { a n } diverges. Solution : We have 0 n ! n n = 1 n 2 n · 3 n . . . n n 1 n . (The last inequality follows since the factor in parentheses is less than one.) More- over, we observe that lim n →∞ 0 = lim n →∞ 1 n = 0. Therefore, by the squeeze theorem, lim n →∞ n ! n n = 0 .
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Math 31B/4 Exam 2 - Page 3 of 7 November 18, 2013 2. (a) (10 points) Find a value of c so that n =0 (1 + c ) - n = 3 2 . ( Hint: Which types of series can we compute?) Solution : Using the formula for the sum of a convergent geometric series, X n =0 (1 + c ) - n = X n =0 1 1 + c n = 1 1 - 1 1+ c , whenever 1 1+ c < 1. Thus, we want a c so that 1 1 - 1 1+ c = 3 2 . Solving for c gives c = 2. (b) (10 points) Let S N be the N th partial sum of X n =1 ln n + 1 n + 2 . Find a formula for S N . Does the series converge or diverge? Solution : We have S N = N X n =1 ln n + 1 n + 2 = ln 2 3 + ln 3 4 + · · · + ln N + 1 N + 2 = ln(2) - ln(3) + ln(3) - ln(4) + · · · + ln( N + 1) - ln( N + 2) = ln(2) - ln( N + 2) . That is, the series telescopes! We observe that lim N →∞ S N = lim N →∞ ln(2) - ln( N + 2) = -∞ . Thus, the series diverges.
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Math 31B/4 Exam 2 - Page 4 of 7 November 18, 2013 3. (a) (10 points) Let f ( x ) = cos( x 2 ). Find a constant K 2 so that | f 00 ( x ) | ≤ K 2 for all 0 x 2. Solution : We have f 0 ( x ) = - 2 x sin( x 2 ) , and f 00 ( x ) = - 2 sin( x 2 ) - 4 x 2 cos( x 2 ) .
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