The series $\sum_{n=1}^{+\infty}\frac{1}{1^2+2^2+\cdots+n^2}.$

How to justify the convergence and calculate the sum of the series: $$\sum_{n=1}^{+\infty}\frac{1}{1^2+2^2+\cdots+n^2}.$$

Solution 1:

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

Solution 2:

For the convergence use a comparison with another sum.

Hint: $$\sum_{i=1}^n i^2 =\frac{n (n+1) (2n+1)}{6}$$ and use partial fraction decomposition.

Since you know that the convergence is absolute, you can change the summation order. (And that is important here).

Maybe another hint is $$\sum_{i=1}^\infty (-1)^i \frac{1}{i}=-\ln(2)$$ This is a result from the Taylor series of the logarithm