# 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: $$\sum_{n=1}^{+\infty}\frac{1}{1^2+2^2+\cdots+n^2}.$$

• Can you compare it to something else that you know converges? In particular, is there something bigger than it that converges? Also, the singular of "series" is still "series". Feb 28 '13 at 21:28
• compare it with $\sum_{n=1}^\infty \frac{1}{n(n+1)}$
– user59671
Feb 28 '13 at 21:29

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

• great thanks for the hint.
– user45099
Feb 28 '13 at 21:53
• Okay, how doesn't this converge to zero? The denominator grows without bound! Mar 2 '13 at 14:47
• @PyRulez The sequence of terms in the sum converges to 0. The series doesn't. Dec 5 '14 at 17:23

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