Does $\sum_{n=2}^\infty \frac{n^2-4}{(n-1)^2(n+3)^2} $ converge?
I used the integral test and found that it does, but it was a bit cumbersome. Is there an easier way?
Thanks
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Do you know limit comparison test? – May 04 '16 at 08:27
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3The series is comparable to $\sum_{n = 2}^{\infty} \frac{1}{n^2}$, which converges. – Travis Willse May 04 '16 at 08:27
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https://en.wikipedia.org/wiki/Limit_comparison_test – May 04 '16 at 08:28
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@Did: Wait, nvm. I see what you're saying.... I can't edit it I will just delete it. – Dair May 04 '16 at 08:34
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Didn't think to use limit comparison - seems obvious now. Thanks everyone! – user336886 May 04 '16 at 09:04
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You can look for a bound. For example $$0\le\frac{n^2-4}{(n-1)^2(n+3)^2}<\frac{n^2}{(n-1)^2n^2}=\frac{1}{(n-1)^2}$$ And it is simple to demonstrate that $$S=\sum_{n=2}^{\infty} \frac{1}{(n-1)^2}$$ converges, and the sum will be bounded between $$0\le\sum_{n=2}^\infty \frac{n^2-4}{(n-1)^2(n+3)^2}\le S$$
user3209698
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The easiest is to use equivalence, since it is a series with positive terms: $$\frac{n^2-4}{(n-1)^2(n+3)^2}\sim_\infty\frac{n^2}{n^2\cdot n^2}=\frac1{n^2}, $$ which converge.
Bernard
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Thanks, that's a great way to do it, but I'm not sure I'm allowed to use it, since it isn't mentioned in the textbook. – user336886 May 04 '16 at 08:45
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If I may give my opinion, it should. That's one of the most efficient ways, throwing off all irrelevant details. – Bernard May 04 '16 at 08:47
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I certainly agree. But I guess the limit comparison test would be the next best thing? – user336886 May 04 '16 at 08:54
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It is the limit comparison test with $c=1$ (and using some basic results on equivalence, which shorten the proof). – Bernard May 04 '16 at 08:58
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By partial fraction decomposition, $$\begin{eqnarray*}\sum_{n\geq 3}\frac{n^2-2n-3}{(n-2)^2 (n+2)^2}&=&\frac{1}{32}\sum_{n\geq 3}\left(\color{blue}{\frac{7}{x-2}-\frac{7}{x+2}}+\color{red}{\frac{10}{(x+2)^2}-\frac{6}{(x-2)^2}}\right)\\&=&\frac{1}{32}\left[\color{blue}{\frac{175}{12}}\color{red}{-6\,\zeta(2)+10\left(\zeta(2)-\frac{1}{1^2}-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}\right)}\right]\\&=&\frac{1}{32}\left[\frac{25}{72}+4\,\zeta(2)\right]=\color{purple}{\frac{25}{2304}+\frac{\pi^2}{48}}.\end{eqnarray*}$$ The blue sum is a telescopic sum, the red sum is a well-known sum.
Jack D'Aurizio
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