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!
On
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 ?
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?