# 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". Commented Feb 28, 2013 at 21:28
• compare it with $\sum_{n=1}^\infty \frac{1}{n(n+1)}$
– user59671
Commented Feb 28, 2013 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
Commented Feb 28, 2013 at 21:53
• Okay, how doesn't this converge to zero? The denominator grows without bound! Commented Mar 2, 2013 at 14:47
• @PyRulez The sequence of terms in the sum converges to 0. The series doesn't. Commented Dec 5, 2014 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