I am trying to approximate $\sum \frac{1}{n^2}$ with telescopic series. So far, I've understood up to this: $$\begin{eqnarray*} \sum_{n\geq m}\frac{1}{n^2}&=&\sum_{n\geq m}\left(\frac{1}{n}-\frac{1}{(n+1)}\right)+\frac{1}{2}\sum_{n\geq m}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\\&+&\frac{1}{6}\sum_{n\geq m}\left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)-\frac{1}{6}\sum_{n\geq m}\frac{1}{n^3(n+1)^3}\end{eqnarray*} $$ I am told that the last term, $-\displaystyle\frac{1}{6}\sum_{n\geq m}\frac{1}{n^3(n+1)^3}$, can be represented as $C\sum(\frac{C}{n^5}-\frac{C}{(n+1)^5})$ plus sum other higher order term, where $C$ is a just a number, different in each appearance.
Questions
How do we make this cubic into a telescopic sum of quintics?
And after we do that, how do we keep going and reating more and more telescopic terms, like $C\sum(\frac{C}{n^7}-\frac{C}{(n+1)^7})$ or something?
$$\frac{1}{n^3(n+1)^3}=\frac{1}{n^6}-\frac{3}{n^7}+\frac{6}{n^8}+\ldots \tag{1}$$ $$\frac{1}{n^5}-\frac{1}{(n+1)^5} = \frac{5}{n^6}-\frac{15}{n^7}+\frac{35}{n^8}\ldots\tag{2} $$ hence $$ \frac{1}{n^3(n+1)^3}-\color{blue}{\frac{1}{5}}\left(\frac{1}{n^5}-\frac{1}{(n+1)^5}\right) = -\frac{1}{n^8}+\ldots \tag{3} $$ where the LHS equals $$ -\frac{1+5 n+5 n^2}{5 n^5 (1+n)^5} \tag{4}$$ which can be approximated by something of the form $\frac{D}{n^7}-\frac{D}{(n+1)^7}$
by just repeating the steps $(1),(2),(3)$.
I am quite fond of the celebrity my proof of Stirling's approximation is acquiring. I got that idea while writing the first section of these notes, before discovering such approach was already well-known in the literature.