<s>[INST] Problem with Summation of series Question: What is the value of $$\frac{1}{3^2+1}+\frac{1}{4^2+2}+\frac{1}{5^2+3} ...$$ up to infinite terms?
Answer: $\frac{13}{36}$
My Approach:
I first find out the general term ($T_n$)$${T_n}=\frac{1}{(n+2)^2+n}=\frac{1}{n^2+5n+4}=\frac{1}{(n+4)(n+1)}=\frac{1}{3}\left(\frac{1}{n+1}-\frac{1}{n+4}\right)$$
Using this, I get,$$T_1=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}\right)$$ $$T_2=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}\right)$$ $$T_3=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}\right)$$
I notice right away that the series does not condense into a telescopic series. How do I proceed further?
 [/INST] You were so close of solving it!
The final step is of the form
$$\frac{1}{3}\sum_{n=1}^\infty{\frac{1}{n+1}-\frac{1}{n+4}}$$
Writing down the series
$$\require{cancel} \frac{1}{3}\left[\left(\frac{1}{2}-\cancel{\frac{1}{5}}\right)+\left(\frac{1}{3}-\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{7})\right)+\left(\cancel{\frac{1}{5}}-\frac{1}{8})\right)\right]$$
So we get the first, second and third positive term and the last three, since we are evaluating it up to infinity, these last ones will be zero and we end up with
$$\frac{1}{3}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)=\frac{1}{3}\left(\frac{13}{12}\right)=\boxed{\frac{13}{36}}$$
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