<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 have solved it already. Recognise, that you can factor out $\frac 13$ from all terms. You have an infinite terms of +/- fractions, from which the first positive three ($\frac 12$, $\frac 13$, $\frac 14$) remains in the sum, every other is cancelled out by a negative counterpart with a 3 step gap. Therefore the end result is $\frac 13 (\frac 12 + \frac 13 + \frac 14)$.
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