142p23s11solution[1]

B n2 n2n1 n2 so given series converges by comparison

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Unformatted text preview: ly the improper integral 1∞ xe−x dx converges. ∞ −x dx = 2/e < ∞ (integrate by parts and use L’Hopital’s rule) so series 1 xe converges. 3.3 Determine if the following series converge or diverge. Give your reasoning using complete sentences. a) b) ∞ ln n 2 n=1 n ∞ n! n=1 (n + 2)! MTH 142 Exam 3 Spr 2011 Practice Problem Solutions 2 a) Series converges if and only the improper integral 1∞ ln2x dx converges. x Integrate by parts and use L’Hopital’s rule to see that this integral converges to 1 so series converges. Alternately, ln x < x1/2 when x is large since by ln L’Hopital’s rule limx→∞ x1/x = 0, so ln2x ≤ x31/2 so 1∞ ln2x dx is convergent by 2 x x comparison with the integral in problem 2a) above 1 1 n! b) (n+2)! = (n+2)(n+1) < n2 so given series converges by comparison with p-series with p = 2 which is convergent. 3.4 For each of the following items a) and b) choose a correct conclusion and reason from among the choices (R), (C), (I) below and provide supporting computation. For example, if you choose (R) calcul...
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