# Find the exact value of the infinite series

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!

• Please use MathJax YOu'll get more help if your questions are easy to read. – saulspatz Jan 28 '19 at 18:53

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?
• @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 – G Cab Jan 28 '19 at 22:56
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 ?