30
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ Do you want to compute its value or only to prove that it converges? $\endgroup$
    – ajotatxe
    Nov 5 '14 at 20:40
59
$\begingroup$

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.

$\endgroup$
20
  • 5
    $\begingroup$ ABSOLUTELY AWESOME!!! (+1) $\endgroup$ Nov 5 '14 at 21:51
  • 1
    $\begingroup$ Very clever. $\;$ $\endgroup$ Nov 6 '14 at 11:11
  • 3
    $\begingroup$ @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 $S_{2m} = \frac{4}{3}\sum\limits_{j=0}^{m}\binom{4m-2j-1}{2m-1}\zeta(2j)\zeta(6m-2j)$ .. $\endgroup$
    – r9m
    Nov 10 '14 at 5:17
  • 3
    $\begingroup$ @hypergeometric This paper contains an elementary derivation of a class of Euler Double summation !! $\endgroup$
    – r9m
    Nov 10 '14 at 7:12
  • 2
    $\begingroup$ 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) $\endgroup$
    – tired
    Feb 13 '17 at 9:22
12
+100
$\begingroup$

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 $n = 1$,

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

$\endgroup$
1
  • 2
    $\begingroup$ Nice use of the Fourier series of the Bernoulli polynomials. (+1) $\endgroup$
    – Mark Viola
    Feb 1 '17 at 19:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.