When the series does converge, it is just the geometric seriesXn=0x-45n=11-x-45=15-(x-4)5=59-x,so when it converges the series converges to the functionf(x) =59-x.17. Does the series∞Xn=1(-1)n(n2+n3)n4+ 1converge absolutely, converge conditionally, or diverge? Explain your answer.Answer:The series of absolute values∞Xn=1(-1)n(n2+n3)n4+ 1=∞Xn=1n2+n3n4+ 1should behave similarly to∑n3n4, so do a limit comparison to this series:limn→∞n2+n3n4n3n4= limn→∞n2+n3n4·n4n3= limn→∞n2+n3n3= 1.Therefore, since∑n3n4=∑1ndiverges, the Limit Comparison Test says that the series of absolutevalues diverges as well.However, the given series satisfies the hypotheses of the Alternating Series Test, so it converges. There-fore, since the series converges but the series of absolute values diverges, we conclude that the seriesconverges conditionally.18. Does the series∞Xn=1(2n)!2n(n!)2converge or diverge? Explain your answer.Answer:Use the Ratio Test:limn→∞(2(n+1))!2n+1((n+1)!)2(2n)!2n(n!)2= limn→∞(2n+ 2)!2n+1((n+ 1)!)2·2n(n!)1(2n)!= limn→∞(2n+ 2)(2n+ 1)2·(n!)2((n+ 1)n!)2= limn→∞(2n+ 2)(2n+ 1)2·1(n+ 1)2.Since (2n+ 2)(2n+ 1) = 4n2+ 6n+ 2 and since (n+ 1)2=n2+ 2n+ 1, the above limit is equal tolimn→∞4n2+ 6n+ 22(n2+ 2n+ 1)=42= 2.Since 2>1, the Ratio Test implies that the given series diverges.7
19. Does the series∞Xn=2(-1)nnln(n)converge absolutely, converge conditionally, or diverge? Explain your answer.Answer:The series of absolute values∞Xn=2(-1)nnln(n)=∞Xn=21nln(n)diverges (see HW #12, Problem 3 for a proof of this). However, the series satisfies the hypotheses ofthe Alternating Series Test and hence converges, so we see that it converges conditionally.20. For what values ofpdoes the series∞Xn=01(n2+ 1)pconverge? Explain your answer.Answer:For largen, the +1 will barely contribute, so do a limit comparison with the series∑1(n2)p:limn→∞1(n2+1)p1(n2)p= limn→∞1(n2+ 1)p·(n2)p1= limn→∞n2n2+ 1p= 1p= 1,since limn→∞n2n2+1= 1.Therefore, the given series and the series∑1(n2)p=∑1n2pwill either both converge or both diverge.Since∑1n2pconverges for 2p >1 and diverges otherwise, we see that the given series converges when2p >1, which is to say whenp >12.21. What is the interval of convergence of the following power series? Explain your answer.∞Xn=13n(x-2)nn2.Answer:Start by applying the Ratio Test to the series of absolute values:limn→∞3n+1(x-2)n+1(n+1)23n(x-2)nn2= limn→∞3n+1|x-2|n+1(n+ 1)2·n23n|x-2|n= limn→∞3|x-2| ·n2(n+ 1)2= 3|x-2|.Therefore, the ratio test implies that the given series converges absolutely when 3|x-2|<1, meaningwhen|x-2|<13.