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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. This happens when 3 p>1,which is to say whenp<2. So the given series converges whenp< would like to estimate the sum of the series n=11n4+3by using the sum of the first ten terms.

9. Consider the sequence defined by a n = (−1)n+n (−1)n−nDoes this sequence converge and, if it does, to what limit? Answer: Dividing numerator and denominator by n, we have that

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