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I'm calculating this double sum: $$ \sum _{m=1}^{\infty } \sum _{k=0}^{\infty } \frac{(-1)^m}{(2 k+1)^2+m^2} $$ I know the answer is $$ \frac{ \pi \log (2)}{16}-\frac{\pi ^2}{16} $$ which can be verified by numerical calculations. I used the Taylor expansions of $log (1+x)$ and $arcsin (x)$ at x=1 to replace $log (2)$ and $\pi$. I got $$ \sum _{m=1}^{\infty } \sum _{k=0}^{\infty } \frac{(2 m+1) (-1)^{k+m}}{4 m(2 k+1) (2 m-1)} $$ I tried to play with the dummy variables but failed. Any idea?

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    $\begingroup$ How do you know the sum is $\pi \log(2)/16 - \pi^{2}/16$? Also, why are you interested in summing this series? $\endgroup$ – Jeremy Rouse Dec 8 '14 at 15:12
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    $\begingroup$ It comes from the probelm 3.48 in Griffiths' <Introduction to Electrodynamics>, 3rd edition. There he met a sum of series $\sum_{n\ odd} \frac{1}{n*sinh (n\pi)}$, which he stated he couldn't find the analytic answer but numerically he got this log and $\pi$ stuff. I'm trying to solve it and I've found it can be converted to the problem of this double sum. $\endgroup$ – tcya Dec 8 '14 at 16:07
  • $\begingroup$ The problem is that your double series in not absolutely convergent. So one cannot interchange the summation, and has to be very careful with what the outer summation means. Stein and Sharkachi textbook on complex analysis addresses this issue and I suppose that a proof can be found there. $\endgroup$ – Alexandre Eremenko Dec 8 '14 at 21:58
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    $\begingroup$ I'll try to write something more detailed later, but this reminds me of another sum I wrote about on math.SE where we had a double sum of one over a quadratic with an alternating sign math.stackexchange.com/questions/426325/… $\endgroup$ – David E Speyer Dec 9 '14 at 17:01
  • $\begingroup$ See also: math.stackexchange.com/q/1117583/37122 $\endgroup$ – Benjamin Dickman Jan 25 '15 at 5:17
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Consider for any real number $s>1$ the double series $$ \sum_{m=1}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^m}{((2k+1)^2+m^2)^s}. \tag{1} $$ This double series converges absolutely. Moreover for any given $m$, the inner series over $k$ converges to some $C(m,s)$, say, which for any given $s \ge 1$ is a monotone decreasing function of $m$. Combined with (the idea behind the) alternating series test, it is not hard to justify that the series in (1) tends to our desired sum as $s\to 1^+$.

Now let $R(n)$ denote the number of ways of writing $n$ as $a^2+b^2$ with $a$ and $b$ integers; note $R(1)=4$ since $1=1^2+0^2= (-1)^2+0^2=0^2+1^2=0^2+(-1)^2$. It is well known that $$ \sum_{n=1}^{\infty} \frac{R(n)}{n^s} = 4\zeta(s) L(s,\chi_{-4}), $$ and that $$ \sum_{n=1, \text{n odd}}^{\infty} \frac{R(n)}{n^s} = 4 \Big(1-\frac{1}{2^s}\Big) \zeta(s)L(s,\chi_{-4}). $$ Here $L(s,\chi_{-4})=1/1^s-1/3^s+1/5^s-1/7^s+\ldots$ is the Dirichlet $L$-function for the character $\pmod 4$.

Now let us return to the sum in (1). Write $(2k+1)^2+m^2=n$. If $n$ is odd, then $m$ is necessarily even, and this number is counted in (1) a total of $R(n)/8$ times; the only exceptions are when $n$ is an odd square, where the solutions $(2k+1)^2+0^2$ are not counted. So the contribution of the odd numbers $n$ to (1) is $$ \frac{1}{8} \sum_{n \text{odd} } \frac{R(n)}{n^s} - \frac{1}{2} \sum_{k=0}^{\infty} \frac{1}{(2k+1)^{2s}}= \frac{1}{2} \Big(1-\frac{1}{2^s}\Big) \zeta(s)L(s,\chi_{-4}) - \frac 12 \Big(1-\frac{1}{2^{2s}}\Big) \zeta(2s). $$

Now consider the contribution of $n\equiv 2\pmod 4$ to (1). These are the terms with $m$ odd, and they appear with sign $-1$. Here each $n$ appears $R(n)/4$ times. So these terms give $$ -\frac{1}{4} \sum_{n\equiv 2\pmod 4} \frac{R(n)}{n^s} = -\frac{1}{4} \frac{1}{2^s} \sum_{\ell \text{ odd} }\frac{R(\ell)}{\ell^s} = -\frac{1}{2^{s}}\Big(1-\frac{1}{2^s}\Big)\zeta(s) L(s,\chi_{-4}), $$ upon writing $n=2\ell$ with $\ell$ odd, and using here $R(n)=R(\ell)$.

Thus our sum in (1) equals $$ \Big(\frac12 -\frac{1}{2^s}\Big) \Big(1-\frac{1}{2^s}\Big) \zeta(s) L(s,\chi_{-4}) - \frac 12 \Big(1-\frac{1}{2^{2s}}\Big) \zeta(2s). $$ Now let $s\to 1^+$ and use $\zeta(2)=\pi^2/6$, $L(1,\chi_{-4})=\pi/4$ (Gregory's formula), and $(1/2-1/2^s) \zeta(s) \to (\log 2)/2$ (zeta has a simple pole with residue $1$ at $1$). The claimed identity follows.

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  • $\begingroup$ Thank you. The answer goes much deeper than I expected. It's actually a special case of Thompson-Lampard theorem in electrodynamics. That's why Griffiths knew the answer though he couldn't prove it for this specific case. Maybe by making other special cases we can get some results for similar infinite series, but both the theorem and the Dirichlet L-function are beyond my current knowledge. If interested, you may refer to the solutions here link for 3.47 and 3.48a for Thompson-Lampard theorem. $\endgroup$ – tcya Dec 11 '14 at 11:40
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Here is a proof based on Hachino's idea. We have agreed it's enough to prove that $$S=\sum_{m=1}^\infty\frac{(-1)^m\tanh(m\pi /2)}{m}=\frac{\log 2-\pi}{4}. $$ Plug in the expansion $$\tanh(m\pi /2)=\frac{1-e^{-m\pi }}{1+e^{-m\pi }}=1+2\sum_{n=1}^\infty(-1)^ne^{-nm\pi }. $$ Changing order of summation and using the Taylor series of $\log(1+x)$ gives $$S=-\log 2+2\sum_{n=1}^\infty(-1)^{n+1}\log(1+e^{-n\pi })=\log\left(\frac{\prod_{n=1}^\infty(1+e^{-(2n-1)\pi})^2}{2\prod_{n=1}^\infty(1+e^{-2n\pi})^2}\right).$$ The infinite products can be recognized as theta values. Recall that $$\theta_2(z,q)=2q^{1/4}\cos z\prod_{n=1}^\infty(1-q^{2n})(1+2q^{2n}\cos 2z+q^{4n}), $$ $$\theta_3(z,q)=\prod_{n=1}^\infty(1-q^{2n})(1+2q^{2n-1}\cos 2z+q^{4n-2}). $$ There are lots of conflicting notations for theta functions; I use the conventions of Whittaker and Watson, A course in modern analysis. It follows that $$S=\log\left(e^{-\pi/4}\frac{\theta_3(0,e^{-\pi})}{\theta_2(0,e^{-\pi})}\right). $$ It remains to show that $$\frac{\theta_3(0,e^{-\pi})}{\theta_2(0,e^{-\pi})}=2^{1/4}. $$ This is clear from generalities on elliptic functions. Again in the notation of Whittaker and Watson, the modulus $k$ is given by $$k=\frac{\theta_2(0,q)^2}{\theta_3(0,q)^2}. $$ The value $q=e^{-\pi}$ corresponds to $k=k'$, where $k'$ is the complementary modulus defined by $k^2+(k')^2=1$. Thus, $k=1/\sqrt 2$, which gives the desired result.

If you had a tough childhood, you might worry that I changed the order of summation in series that are not absolutely convergent. You might want to plug in a convergence factor $x^m$, repeat the argument above and prove that both sides of the resulting identity are left continuous at $x=1$.

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  • $\begingroup$ Thank you. The answer goes much deeper than I expected. It's actually a special case of Thompson-Lampard theorem in electrodynamics. That's why Griffiths knew the answer though he couldn't prove it for this specific case. Maybe by making other special cases we can get some results for similar infinite series, but both the theorem and the theta function are beyond my current knowledge. If interested, you may refer to the solutions here link for 3.47 and 3.48a for Thompson-Lampard theorem. $\endgroup$ – tcya Dec 11 '14 at 11:40
  • $\begingroup$ Sounds very interesting. I'm not surprised that elliptic functions are useful for this kind of physical problem. The formula for the potential in your link also seems directly related to elliptic functions. $\endgroup$ – Hjalmar Rosengren Dec 11 '14 at 12:34
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Well, using WolframAlpha, one has the identity

\begin{equation} \sum_{k=0}^{\infty} \frac{1}{k^2 + a^2} = \frac{\pi a \coth (\pi a) + 1}{2a^2} \end{equation}

which, after separating the sum into odd/even contributions and factorizing the terms, leads to :

Edit : corrected after Hjalmar's remark

\begin{equation} \sum_{k=0}^\infty\frac 1{(2k+1)^2+m^2}=\frac{\pi\tanh(\pi m/2)}{4m} \end{equation}

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    $\begingroup$ I don't think that $(-1)^m/\coth(\pi m/2)\rightarrow 0$ as $m\rightarrow\infty$? The denominator tends to 1. $\endgroup$ – NAME_IN_CAPS Dec 8 '14 at 18:22
  • $\begingroup$ Exactly. The required answer can be obtained only if the sum of this divergent series is given a precise meaning. For this see the Stein and Sharkachi textbook on Complex analysis. $\endgroup$ – Alexandre Eremenko Dec 8 '14 at 22:01
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    $\begingroup$ You made a mistake when separating the sum into even and odd parts. The correct result is $$\sum_{k=0}^\infty\frac 1{(2k+1)^2+m^2}=\frac{\pi\tanh(\pi m/2)}{4m}.$$ This simplifies the double series to the single series $$\sum_{m=1}^\infty\frac{(-1)^m\tanh(\pi m/2)}{m}=\frac{\log(2)-\pi}4. $$ This is convergent (but not absolutely convergent). I am not sure about the simplest way to prove this summation. Probably you can start from the Fourier expansion of an elliptic function and plug in special values, but I am sure there are more elementary ways. $\endgroup$ – Hjalmar Rosengren Dec 9 '14 at 8:39
  • $\begingroup$ Oops, sorry, should've checked twice rather than once. I tried to expand the $\tanh$ using the series of $\frac{1}{1+x}$, which led to the appearance of the $\ln$, but then the result is not obviously equal to $\frac{\log(2)-\pi}4$. $\endgroup$ – Hachino Dec 9 '14 at 9:36
  • $\begingroup$ Hi @HjalmarRosengren, your last step is exactly what I'm trying to prove and idk how. After playing with it I got the double sum in my question. I guess maybe natural numbers are easier to handle so I didn't ask for the original sum of $tanh$. $\endgroup$ – tcya Dec 9 '14 at 14:57
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So I think this can be done, in two steps:

(1) understand sums of the 1/Q(k, m) for certain integral quadratic forms Q; and

(2) express this sum as a linear combination of such sums.

The first part is classical number theory, while the second part is elementary, though in general the answers in (1) will come from double sums over all pairs of integers (positive and negative), obviously omitting (0,0).

The simplest case for (1) is where Q is the square of the distance of the lattice point (k,m) to the origin. This actually diverges ... so while this is all perfectly standard number theory, you need to be taking care that the linear combination cancels the divergences. So the Dedekind zeta function of the Gaussian integers, and related functions, should be invoked.

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