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Math 115 Exam #1 Practice Problems

Math 115 Exam #1 Practice ProblemsFor each of the following, say whether it converges or diverges and explain n=1n3n5+3 Answer:Notice thatn3n5+ 3<n3n5=1n2for alln. Therefore, since 1n2converges (it s ap-series withp= 2>1), the series n3n5+3alsoconverges by the comparison n=13n4n+4 Answer:Notice that3n4n+ 4<3n4n=(34)nfor alln. Therefore, since (34)nconverges (it s a geometric series withr=34<1), the series 3n4n+4also converges by the comparison n=1n2nAnswer:Using the Root Test:limn n n2n = limn n nn 2n= limn n n2= the limit is less than 1, the Root Test says that the series converges what values ofpdoes the series n=1np2+n3converge?Answer:Doing a limit comparison to 1n3 p, I see thatlimn np2+n31n3 p= limn n32 +n3= , the series converges if and only if the series 1n3 pconverges.

Math 115 Exam #1 Practice Problems For each of the following, say whether it converges or diverges and explain why. 1. P ∞ n=1 n3 5+3 Answer: Notice that n3 n5 +3 < n3 n5 = 1 n2 for all n. Therefore, since P 1 n2 converges (it’s a p-series with p = 2 > 1), the series P n3 n5+3 also converges by the comparison test. 2. P ∞ n=1 3n 4n+4 ...

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