Sum of (arithmetic?) infinite series How the heck do I find the sum of a series like $\sum\limits_{n=3}^\infty\frac{5}{36n^{2}-9}$? I can't seem to convert this to a geometric series and I don't have a finite number of partial sums, so I'm stumped.
 A: Hint: See if the series telescopes if you were to approach via partial fraction decomposition.
A: Notice, we have $$\sum_{n=3}^{\infty}\frac{5}{36n^2-9}$$ $$=\frac{5}{9}\sum_{n=3}^{\infty}\frac{1}{4n^2-1}$$ 
$$=\frac{5}{9}\sum_{n=3}^{\infty}T_n$$
$$\sum_{n=3}^{\infty}\frac{5}{36n^2-9}=\frac{5}{9}\lim_{n\to \infty}\sum_{n=3}^{n}T_n\tag 1$$
Where, $T_n$ is nth term of the series 
given as
$$T_n=\frac{1}{4n^2-1}= \frac{1}{(2n-1)(2n+1)}$$ $$=\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)$$ Now, setting $n=3, 4, 5, \dots n$, we get  $$T_1=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}\right)$$ $$T_2=\frac{1}{2}\left(\frac{1}{7}-\frac{1}{9}\right)$$ $$T_3=\frac{1}{2}\left(\frac{1}{9}-\frac{1}{11}\right)$$ $$..................$$
$$...................$$ 
$$T_{n-1}=\frac{1}{2}\left(\frac{1}{2n-3}-\frac{1}{2n-1}\right)$$
$$T_{n}=\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)$$
Now, adding all the terms column-wise we get sum of all $(n-2)$ terms as follows $$\sum_{n=3}^{n} T_n=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{2n+1}\right)$$
Now, taking limit as $n\to \infty$, we get $$\lim_{n\to \infty}\sum_{n=3}^{n} T_n=\lim_{n\to \infty}\frac{1}{2}\left(\frac{1}{5}-\frac{1}{2n+1}\right)$$ $$=\frac{1}{2}\lim_{n\to \infty}\left(\frac{1}{5}-\frac{1}{2n+1}\right)$$ $$=\frac{1}{2}\left(\frac{1}{5}-0\right)=\frac{1}{10}$$ Now, setting the above value in (1), we get
$$\sum_{n=3}^{\infty}\frac{5}{36n^2-9}=\frac{5}{9}\times \frac{1}{10}$$ $$=\frac{1}{18}$$
A: $$\frac{5}{9}\sum \frac{1}{4x^2-1}$$
$$=\frac{5}{18}\sum \frac{1}{2x-1}-\frac{1}{2x+1}$$
$$=\frac{5}{18}[ \frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}...]$$
$$=\frac{5}{18}[\frac{1}{5}]$$
$$=\frac{1}{18}$$
