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Find the exact value of the infinite series given by $$S = \frac{1}{(3^2+1)} + \frac{1}{(4^2+2)} + \frac{1}{(5^2+3)} + ... $$

I found the notation of how it would be written with sigma notation: $$\sum_{x=1}^∞ \frac{1}{(x+1)(x+4)}$$

I don't know how to get an exact value for the sum after that.

Help would be appreciated, thanks!

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    $\begingroup$ Please use MathJax YOu'll get more help if your questions are easy to read. $\endgroup$ – saulspatz Jan 28 '19 at 18:53
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Hint: $$\frac{1}{(x+1)(x+4)}=\frac{1}{3}\left(\frac{1}{x+1}-\frac{1}{x+4}\right).$$ Can you see how to compute the telescopic sum?

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  • $\begingroup$ Thanks, but I'm not too sure on what telescopic sums are. $\endgroup$ – ShadyAF Jan 28 '19 at 20:40
  • $\begingroup$ @ShadyAF to see that, just write write down $\frac{1}{n+1}-\frac{1}{n+4}$, for the few first values of $n$, one row for each $n$, and sum $\endgroup$ – G Cab Jan 28 '19 at 22:56
  • $\begingroup$ Hey, thanks! I got it! $\endgroup$ – ShadyAF Jan 29 '19 at 2:07
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You can write your sum as $$ \sum_{n=1}^{+\infty}\frac{1}{n+\left(n+2\right)^2} $$ and $$ \frac{1}{n+\left(n+2\right)^2}=\frac{1}{n^2+5n+4} $$ Then you need to write $$ \frac{1}{n^2+5n+4}=\frac{a}{n-\alpha_1}+\frac{b}{n-\alpha_2} $$ with $\alpha_1, \alpha_2$ being the two roots of $n^2+5n+4$. Can you take it from there ?

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