Determine $s_{10}$ for $\sum_{n=1}^{\infty}\frac{1}{n^2}$ Consider the convergent series $$\sum_{n=1}^{\infty}\frac{1}{n^2}$$
To determine $s_{10}$ which is the sum of the first ten terms, the easiest way of course is to add them up.
However is there another way to figure out $s_{10}$?
 A: By creative telescoping
$$\begin{eqnarray*} \sum_{n\geq m}\frac{1}{n^2}&=&\sum_{n\geq m}\left(\frac{1}{n}-\frac{1}{(n+1)}\right)+\frac{1}{2}\sum_{n\geq m}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\\&+&\frac{1}{6}\sum_{n\geq m}\left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)-\frac{1}{6}\sum_{n\geq m}\frac{1}{n^3(n+1)^3}\tag{1}\end{eqnarray*} $$
hence by recalling that $\zeta(2)=\frac{\pi^2}{6}$ and plugging in $m=11$ in $(1)$ we get:
$$ H_{10}^{(2)} = \frac{\pi^2}{6}-\frac{1}{11}-\frac{1}{2\cdot 11^2}-\frac{1}{6\cdot 11^3}+\frac{1}{6}\sum_{n\geq 11}\frac{1}{n^3(n+1)^3}\tag{2} $$
hence $\frac{\pi^2}{6}-\frac{1}{11}-\frac{1}{2\cdot 11^2}-\frac{1}{6\cdot 11^3}$ is an approximation of $H_{10}^{(2)}$ with an error $\leq 10^{-6}$.
By Wolstenholme's theorem we know that $11$ is a divisor of the numerator of $H_{10}^{(2)}$ and the denominator of $H_{10}^{(2)}$ is clearly a divisor of $2^6\cdot 3^4\cdot 5^2\cdot 7^2$. These facts allow to turn the previous approximation into an exact evaluation:
$$ H_{10}^{(2)}=\frac{1968329}{1270080}\tag{3}$$
but I wonder why a reasonable person should follow this approach, instead of just adding ten terms of the form $\frac{1}{n^2}$.
