{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

142p23s11solution[1]

# B n2 n2n1 n2 so given series converges by comparison

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}