How can I prove that $\sum\limits^\infty_{n=0}\frac{3n^2+2} {n \cdot (5n-1)^2}$ is convergent? I failed doing it with the comparison test and root test aswell as fraction test seem far too complicated to me. Is there an easy way?
Prove that $\sum\limits^\infty_{n=0}\frac{3n^2+2} {n \cdot (5n-1)^2}$ is convergent
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$\begingroup$
limits
convergence
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0Is this convergence of a sequence? If so, divide the top and bottom by $n^3$, and note that $\frac 1n$ goes to zero as $n$ increases. – 2017-01-11
3 Answers
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Fill in details:
$$\frac{3n^2+2}{n(5n-1)^2}\ge\frac{3n^2}{n\cdot25n^2}=\frac3{25}\cdot\frac1{n}$$
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0to conclude that the series diverges, right? – 2017-01-11
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0Ohh okay I see. Thank you so it doesn't even converge... That's why I am not able to prove it. – 2017-01-11
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0@RSerrao Yes, of course... – 2017-01-11
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0@user405981 Indeed. But also pay attention to the fact that you should at least remark the series is a positive one so you can use the comparison test at all... – 2017-01-11
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Using the Comparison Ratio Test: $$\lim_{n\to +\infty}\left(\frac{3n^2+2} {n (5n-1)^2}:\frac{1}{n}\right)=\frac{3}{25}\ne 0\Rightarrow\sum_{n=1}^{+\infty} \frac{3n^2+2} {n (5n-1)^2}\sim \sum_{n=1}^{+\infty} \frac{1} {n}\text{ (divergent).}$$
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See Limit Comparison Test. As $a_n$ is of order $1/n$ the series will diverge.
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0I was talking about a series sorry. Edited the post. – 2017-01-11
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0Then the series will not converge as it is of order $1/n$ – 2017-01-11
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0@Sachidanand Prasad And how can I prove that? – 2017-01-11
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0See the answer. – 2017-01-11
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0Why downvote is given? – 2017-01-11