TITLE: The series $\sum_{n=1}^{+\infty}\frac{1}{1^2+2^2+\cdots+n^2}.$
QUESTION [22 upvotes]: How to justify the convergence and calculate the sum of the series:
$$\sum_{n=1}^{+\infty}\frac{1}{1^2+2^2+\cdots+n^2}.$$

REPLY [35 votes]: $$\begin{array}{lcl}
\sum_{n=1}^\infty \frac{1}{1^2+2^2+\cdots+n^2}&=& \sum_{n=1}^\infty\frac{6}{n(n+1)(2n+1)} \\ &=& 6\sum_{n=1}^\infty \frac{1}{2n+1} \left( \frac{1}{n}-\frac{1}{n+1}\right) \\ &=& 12\sum_{n=1}^\infty \frac{1}{2n(2n+1)} -12\sum_{n=1}^\infty \frac{1}{(2n+1)(2n+2)} \\ &=& 12\sum_{n=1}^\infty \left[ \frac{1}{2n}-\frac{1}{2n+1} \right] - 12\sum_{n=1}^\infty \left[ \frac{1}{2n+1}-\frac{1}{2n+2} \right]\\
&=& 12(1-\ln 2)- 12\left(\ln 2-\frac{1}{2}\right)\\ &=& 18-24\ln 2
\end{array}
$$