I ran across another challenging and interesting series, and I am wondering if someone could shed some light on how to evaluate it.

$$ \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{(n^{2}+k^{2})^{2}}=\zeta(2)\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n-1)^{2}}-\zeta(4)$$

This has turned out to be rather challenging. At first glance, I thought it may be somewhere along the lines of the famous $$ \sum_{n=1}^{\infty}\frac{1}{n^{2}+k^{2}}=\frac{\pi}{2k}\coth(\pi k)-\frac{1}{2k^{2}}$$ that is often seen in Complex Analysis.

So, I ran the first sum through Maple and it gave me:

$$ \sum_{n=1}^{\infty} \sum_{k=1}^{\infty}\frac{1}{(n^{2}+k^{2})^{2}}=\frac{{\pi}^{2}}{4}\sum_{n=1}^{\infty}\frac{\coth^2(\pi n)}{n^{2}}+\frac{\pi}{4}\sum_{n=1}^{\infty}\frac{\coth(\pi n)}{n^{3}}-\frac{{\pi}^{2}}{4}\sum_{n=1}^{\infty}\frac{1}{n^{2}}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^{4}}$$

Of course, two of these are the very familiar $\zeta(2), \;\ \zeta(4)$. I managed to evaluate $\displaystyle \sum_{n=1}^{\infty}\frac{\coth(\pi n)}{n^{3}}=\frac{7{\pi}^{3}}{180}$ using Complex Analysis.

The one that has given me the fit is $\displaystyle\sum_{n=1}^{\infty}\frac{\coth^{2}(\pi n)}{n^{2}}$.

This evaluates to $\displaystyle\frac{2}{3}K+\frac{19{\pi}^{2}}{180}$. But, how?.

I tried a Complex Analysis method, but had trouble finding the residues at $ni$, which are the zeroes of $\sinh^{2}(\pi n)$ .

Anyone have any good ideas on how to evaluate the original sum at the top or even just this 'coth-squared' one?. Complex analysis or other wise. I thought maybe a clever use of Fourier would work, but maybe not.

  • 1
    $\begingroup$ It wasn't until I got to the end of the question that I realised "CA" meant "complex analysis"... also I personally think it's unhelpful to put "interesting" in your title – I already assume that your question is interesting, else why would you post it? $\endgroup$ – Ben Millwood Sep 16 '12 at 11:48
  • $\begingroup$ Ha. I thought CA meant Computer Algebra. $\endgroup$ – Gerry Myerson Sep 16 '12 at 11:51
  • $\begingroup$ See meta.math.stackexchange.com/questions/5082/…. $\endgroup$ – joriki Sep 16 '12 at 14:18
  • $\begingroup$ Sorry, I did not realize that was such an issue. $\endgroup$ – Cody Sep 16 '12 at 15:55

A more general form is known : $$\tag{1}f(c)=\sum_{n,m}'\frac 1{(n^2+m^2)^c}=4\,\zeta(c)\,L_{-4}(c)$$ with $\ L_{\sigma}(c):=\sum_{n\ge 1} \,\left(\frac {\sigma}n\right)\,n^{-c}\quad$ (where $\ \displaystyle\left(\frac {\sigma}n\right)$ is the Kronecker symbol and the sum is on all signed integers $n,m$ excluding $n=m=0$ as indicated by ' )

For $\;\sigma=-4\ $ only the odd terms remain with alternating signs : $$\tag{2}L_{-4}(c)=\sum_{n\ge 1} \,\left(\frac {-4}n\right)\,n^{-c}=\sum_{n\ge 1} \frac {(-1)^{n-1}}{(2n-1)^c}=\beta(c)$$ with $\beta$ the Dirichlet beta function so that (distinguishing the values $n=0$ and $m=0$) : $$4\,\zeta(c)\,\beta(c)=\sum_{n,m}'\frac 1{(n^2+m^2)^c}=2\sum_{m\ge 1}\frac 1{(0^2+m^2)^c}+2\sum_{n\ge 1}\frac 1{(n^2+0^2)^c}+4\sum_{n,m\ge 1}\frac 1{(n^2+m^2)^c}$$ and (dividing by $4$) a generalization of your nice result : $$\tag{3}\sum_{n,m\ge 1}\frac 1{(n^2+m^2)^c}=\,\zeta(c)\,\beta(c)-\zeta(2c)$$

The (rather non-trivial) derivation of a further generalization of $(1)$ is provided in Borwein, Bailey and Girgensohn's "Experimentation in Mathematics" (page $167-169$ of my $2004$ edition).

They define :

$$\tag{4}\zeta(a,b,c) = \sum_{n,m}' \frac{n^{2a}\,m^{2b}}{(n^2+m^2)^c}$$

($a,\;b,\;c\;$ are nonnegative integers)
(the sum was for $n,m\ge 0$ but I think that "$\ge 0$" is a typo that doesn't appear page $169$)

The page $168$ of the book contains (I hope that a one page re-transcription will not be a problem, extracts of the book may be viewed at Amazon) :

"The following identity expresses $\zeta(a,0,c)$ by using a Bessel function expansion of the normalized Mellin transform $$M_c(f) = \frac 1{\Gamma(c)} \int_0^\infty f(t)\,t^{c-1}\,dt$$ of the function $$t \to \sum_{n,m} n^{2a}q^{n^2+m^2} =\sum_n n^{2a}q^{n^2} \theta_3(q)$$ with $q=e^{-t}$ after using the theta transform $(2.15)$ $$\theta_3\bigl(e^{\pi t}\bigr)= \sqrt{\frac{\pi}t}\,\theta_3\left(e^{\frac{\pi}t}\right)$$ This leads to the identity $$\tag{5}\zeta(a,0,c)= 2\delta_{0a}\zeta(2c)+ 2\beta\left(c-\frac12,\frac12\right) \zeta(2c-2a-1)+4\sum_{p\ge 1} \sigma_{[2a+1-2c]}(p)\,{\cal E}_c(p)$$ This is valid for real $ac$ with $d=c-a>1$ and presumably provides an analytic continuation of $\zeta(a,0,c)$ for the sum over positive integers. Here $\sigma$ is a divisor function $$\sigma_{[d]}(p) =\sum_{n|p} n^d,$$ and $${\cal E}_c(p)= \frac{\sqrt{\pi}}{\Gamma(c)} 2(\pi p)^{c-1/2}K_{\bigl(c-\frac 12\bigr)}(2\pi p)$$ is derived from the modified Bessel function of half integer order, $K_{\bigl(c-\frac 12\bigr)}$, of the second kind.

When $c=N$ is integer, ${\cal E}_N(p)$ is of the form $$\pi\,e^{-2\pi p} P_N(\pi p)$$ where $P_N$ is a rational polynomial of degree $N-1$ with positive coefficients : $P_1(x)=1,\ P_2(x)=x+1/2,\ P_3(x)=x^2/2+3/4x+3/8,\ P_4(x)=1/6x^3+1/2x^2+5/8x+5/16$.

In general $$P_N(x)=\sum _{k=0}^{N-1}\frac{{{N+k-1}\choose{N-1}}x^{N-1-k}}{(N-k-1)!4^k}$$ This allows one to very efficiently compute $\zeta(a,0,c)$ via $(5)$, using roughly $D/4$ terms for $D$ digits. Note also that for fixed $c$ and variable $a$ only the powers in $\sigma_{[2a+1-2c]}$ vary so most of the computation can be saved.

Then the general integer case follows from $$\tag{6}\zeta(a,b,c)=\sum_{k=0}^e (-1)^{e-k}{{e}\choose{k}}\zeta(a+b-k,0,c-k)$$ where $\ e=\min(a,b)$."

The formula at the beginning was $\zeta(0,0,c)$ and you asked merely for $\zeta(0,0,2)$. I think there is a much simpler answer for the second one (so that the problem remains...) but generalizations are of value too!

Let's add some keywords and references I just found (MathWorld and others) :

Hoping this helped too,

  • $\begingroup$ Wow, thanks Raymond. Yes, help is always appreciated. I must admit, I am not familiar with most of this. I will have to look into it all further. I managed to evaluate the series by summing each of the four series I mentioned in my first post. The tricky one was the $\frac{\coth^{2}(n\pi)}{n^{2}}$. Thanks much. You are always helpful. $\endgroup$ – Cody Sep 18 '12 at 18:34
  • $\begingroup$ Glad it helped @Cody ! For other generalizations see page $19$ of this. Page $25$ has even a sum with three squares as well as page $378$ of this paper ! For theta and Bessel functions DLMF may help. $\endgroup$ – Raymond Manzoni Sep 18 '12 at 22:53
  • $\begingroup$ Thanks for the edit @Lucian! (concerning ^' I think it worked until around one and half year earlier... possibly a Mathjax evolution since I still use the same old safari browser). Cheers, $\endgroup$ – Raymond Manzoni Dec 18 '14 at 22:16

I managed to evaluate the series by beginning with:

$\displaystyle \sum_{k=1}^{\infty}\frac{1}{n^{2}+k^{2}}=\frac{\pi n\coth(\pi n)-1}{2n^{2}}$, and using some known sums/identities.

Then, differentiate, do some manipulating and it works out to what Maple gave.

Then, sum these four series separately.

It works out to $ \frac{-{\pi}^{2}}{4}\zeta(2)+\frac{{\pi}^{2}}{4}\left(\frac{2}{3}K+\frac{19{\pi}^{2}}{180}\right)-\frac{1}{2}\zeta(4)+\frac{\pi}{4}\cdot \frac{7{\pi}^{3}}{180}$

This then simplifies down to $\zeta(2)K-\zeta(4)$. K being the Catalan constant represented by the series in the problem.

The trickiest sum was $\displaystyle \sum_{n=1}^{\infty}\frac{\coth^{2}(\pi n)}{n^{2}}$.

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



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