# An Infinite Double Summation $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}$?

While Solving some integral problem, I encountered the following infinite series:

$$\displaystyle \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}$$

I have tried many methods including partial fractions... I seek help! Please provide hints if you don't have the complete answer.

• Do you want to compute its value or only to prove that it converges? Commented Nov 5, 2014 at 20:40

We have \begin{align} \displaystyle \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2} & = \frac{1}{3}\sum\limits_{n,k=1}^{\infty}\frac{(n+k)^3 - n^3-k^3}{n^3k^3(n+k)^3} \\ &= \frac{1}{3}\sum\limits_{n,k=1}^{\infty} \frac{1}{n^3k^3} - \frac{1}{3}\sum\limits_{n,k=1}^{\infty}\frac{1}{n^3(n+k)^3} - \frac{1}{3}\sum\limits_{n,k=1}^{\infty}\frac{1}{k^3(n+k)^3} \tag{*}\\ &=\frac{1}{3}\sum\limits_{n,k=1}^{\infty} \frac{1}{n^3k^3} - \frac{1}{3}\sum\limits_{n=1}^\infty\sum\limits_{k = n+1}^\infty \frac{1}{n^3k^3} - \frac{1}{3}\sum\limits_{k=1}^\infty\sum\limits_{n=k+1}^\infty \frac{1}{n^3k^3} \\&= \frac{1}{3}\sum\limits_{n,k=1}^{\infty} \frac{1}{n^3k^3} - \frac{1}{3}\sum\limits_{1 \le n < k <\infty} \frac{1}{n^3k^3} - \frac{1}{3}\sum\limits_{1 \le k < n < \infty} \frac{1}{n^3k^3} \\&= \frac{1}{3}\sum\limits_{n=k=1}^\infty \frac{1}{n^3k^3} = \frac{1}{3}\zeta(6) \end{align}

where, in line $$(*)$$ we reindexed the summation with the change of variables $$n+k \mapsto k$$ in the second sum and $$n+k \mapsto n$$ in the third sum.

• ABSOLUTELY AWESOME!!! (+1) Commented Nov 5, 2014 at 21:51
• Very clever. $\;$ Commented Nov 6, 2014 at 11:11
• @hypergeometric ^_^ .. seems we can, denoting the series by $\displaystyle S_m = \sum_{k=0}^\infty \sum_{n=0}^\infty \frac1{n^mk^m(n+k)^m}$, we can show $S_{2m+1} = -4\sum\limits_{j=0}^{m}\binom{4m-2j+1}{2m}\zeta(2j)\zeta(6m-2j+3)$ and $S_{2m} = \frac{4}{3}\sum\limits_{j=0}^{m}\binom{4m-2j-1}{2m-1}\zeta(2j)\zeta(6m-2j)$ ..
– r9m
Commented Nov 10, 2014 at 5:17
• @hypergeometric This paper contains an elementary derivation of a class of Euler Double summation !!
– r9m
Commented Nov 10, 2014 at 7:12
• interestingly we can show that this series also equals $$I=\int_{[0,1]^2}dxdy\frac{\text{Li}^2_2(xy)}{xy}$$ which is terribly difficult to solve directly (and also mathematica didn't diggest it) Commented Feb 13, 2017 at 9:22

We have, $\displaystyle B_{2n}(x) = (-1)^{n-1}\frac{2(2n)!}{(2\pi)^{2n}} \sum\limits_{k=1}^{\infty} \frac{\cos \left(2\pi kx \right)}{k^{2n}}$

Cubing and integrating both sides,

\begin{align*}&\int_0^1 \left((-1)^{n-1}\frac{(2\pi)^{2n}}{2(2n)!}B_{2n}(x)\right)^3\,dx \\&= \int_0^1 \left(\sum\limits_{k=1}^{\infty} \frac{\cos \left(2\pi kx \right)}{k^{2n}}\right)^3\,dx \\&= \sum\limits_{k_1,k_2,k_3 = 1}^{\infty} \int_0^1 \frac{\cos \left(2\pi k_1x \right)\cos \left(2\pi k_2x \right)\cos \left(2\pi k_3x \right)}{k_1^{2n}k_2^{2n}k_3^{2n}}\,dx\\&= \frac{1}{4}\sum\limits_{k_1,k_2,k_3 = 1}^{\infty} \dfrac{\displaystyle \sum\limits_{\{\pm\}} \int_0^1 \cos \left(2\pi (k_1\pm k_2 \pm k_3)x \right)\,dx}{k_1^{2n}k_2^{2n}k_3^{2n}}\\&= \frac{3}{4}\sum\limits_{k_1,k_2= 1}^{\infty} \frac{1}{k_1^{2n}k_2^{2n}(k_1+k_2)^{2n}}\end{align*}

Since, $\displaystyle \int_0^1 \cos \left(2\pi (k_1 + k_2 - k_3)x \right)\,dx = \begin{cases}1 & \text{ when } k_3 = k_1 + k_2\\ 0 &\text{ otherwise }\end{cases}$

The case $n = 1$,

$$\sum\limits_{k_1,k_2= 1}^{\infty} \frac{1}{k_1^{2}k_2^{2}(k_1+k_2)^{2}} = \frac{4\pi^6}{3}\int_0^1 \left(B_2(x)\right)^3\,dx = \frac{1}{3}\frac{\pi^6}{945} = \frac{1}{3}\zeta(6)$$

• Nice use of the Fourier series of the Bernoulli polynomials. (+1) Commented Feb 1, 2017 at 19:21