I'm trying to compute this series.
$$S=\sum_{k=4}^\infty \dfrac{9}{(k-1)^2} - \dfrac{9}{(k+2)^2}$$
I have found the answer using a "shortcut" aka the Basel sum... $$S=\sum_{k=4}^\infty \dfrac{9}{(k-1)^2} - \dfrac{9}{(k+2)^2} = \sum_{k=4}^\infty \dfrac{9}{(k-1)^2} - \sum_{k=4}^\infty \dfrac{9}{(k+2)^2}$$
Let $k-1=m$ and $k+2=n$, then
$$S=\sum_{m=3}^\infty \dfrac{9}{m^2} - \sum_{n=6}^\infty \dfrac{9}{n^2} = 9\left[\left(\dfrac{\pi^2}{6}-\dfrac{5}{4}\right) - \left(\dfrac{\pi^2}{6}-\dfrac{5269}{3600}\right)\right]= \dfrac{769}{400}$$
This appears to be a telescoping series, so I'm looking for an algebraic approach.
Wolfram Alpha computed the partial sum as
$$S_n =\dfrac{769 n^6 + 4614 n^5 - 803 n^4 - 33972 n^3 - 61724 n^2 - 43200 n - 14400}{400 n^2 (n + 1)^2 (n + 2)^2}$$
Can someone provide the details or outline the steps to arrive at $S_n$?
$$\sum_{m=3}^n\dfrac9{m^2}-\sum_{m=3}^n\dfrac9{(m+3)^2}$$
$$=9\left(\dfrac19+\dfrac1{16}+\dfrac1{25}-\dfrac1{(n+1)^2}-\dfrac1{(n+2)^2}-\dfrac1{(n+3)^2}\right)$$