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Consider the following double sum $$ S = \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{1}{a (2n-1)^2 - b (2m-1)^2} \, , $$ where $a$ and $b$ are both positive real numbers given by \begin{align} a &= \frac{1}{2} - \frac{\sqrt{2}}{32} \, , \\ b &= \frac{1}{4} - \frac{3\sqrt{2}}{32} \, . \end{align} It turns out that one of the two sums can readily be calculated and expressed in terms of the tangente function. Specifically, $$ S = \frac{\pi}{4\sqrt{ab}} \sum_{m=1}^\infty \frac{\tan \left( \frac{\pi}{2} \sqrt{\frac{b}{a}} (2m-1) \right)}{2m-1} \, . $$

The latter result does not seem to be further simplified. i was wondering whether someone here could be of help and let me know in case there exists a method to evaluate the sum above. Hints and suggestions welcome.

Thank you

PS From numerical evaluation using computer algebra systems, it seems that the series is convergent. This apparently would not be the case if $b<0$.

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    $\begingroup$ Maybe Poisson summation formula? $\endgroup$ – Angina Seng Jan 15 '19 at 17:28
  • $\begingroup$ @LordSharktheUnknown Thanks for the comments. Could you please be a bit specific. An example would be highly appreciated $\endgroup$ – Daddy Jan 15 '19 at 17:31
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    $\begingroup$ On second thoughts, I believe now that the double sum is divergent. $\endgroup$ – Angina Seng Jan 15 '19 at 18:07
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    $\begingroup$ @LordSharktheUnknown i agree. The divergence looks logarithmic. Thanks for the hint $\endgroup$ – Daddy Jan 16 '19 at 7:25
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    $\begingroup$ $$ S=\frac{\pi}{4\sqrt{ab}}\sum_{m=1}^\infty\frac{\tan\left(\frac{\pi}{2}\sqrt{\color{red}{\frac{b}{a}}}\left(2m-1\right)\right)}{2m-1} $$ $\endgroup$ – Hazem Orabi Jan 24 '19 at 13:11
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Answer: The series diverges when $\sqrt{\frac{b}{a}}$ is irrational, as in your case. This follows from your reduced form of the sum and a few observations:

  1. $\tan{\frac{\pi x}{2}}$ has period 2 and diverges like $\frac{1}{n - x}$ near any odd integer $n$. More precisely, there exists a constant $c_1$ so that for sufficiently small $x$ $$\left|\tan{\frac{\pi (n - x)}{2}}\right| > \left|\frac{c_1}{x}\right|$$ for any odd $n$.
  2. By the Dirichlet approximation theorem, for irrational $\sqrt{\frac{b}{a}}$ there are infinitely many integers $m$ such that $\sqrt{\frac{b}{a}} (2m-1)$ is within distance $\frac{c_2}{2m-1}$ of an odd integer, for some positive constant $c_2$.
  3. Combining the above two remarks, we find that there are infinitely many $m$ such that $$\left| \frac{\tan{\frac{\pi}{2}\sqrt{\frac{b}{a}} (2m-1)}}{2m-1} \right|>\frac{c_1}{c_2}$$

Thus there are infinitely many terms in the sum which are bigger in magnitude than some positive constant, so the sum can't converge.


Remark: If $\sqrt{\frac{b}{a}}$ is some rational number $\frac{p}{q}$, then the convergence depends on the parity of $p$ and $q$. I'll sketch what happens in each case:

  • If $p$ and $q$ are odd, then for some $m$ we find $\sqrt{\frac{b}{a}} (2m-1)$ will be an odd integer and the tangent in the sum will diverge. Thus the sum doesn't converge.
  • If $p$ is odd and $q$ is even, then $\sqrt{\frac{b}{a}} (2m-1)$ is never an integer, and is symmetrically distributed across the period of the tangent. Note that the tangent is an odd function when reflected across any even integer, so for each negative value of the tangent in the sum there is a corresponding equal and opposite positive value. Within each period of the tangent there are a finite number of such pairs, and the contribution of each such pair to the sum goes like $~\frac{1}{m^2}$, which is convergent. (This is exactly analogous to how $\sum \frac{(-1)^n}{n}$ converges conditionally.) Thus the sum will (conditionally) converge.
  • If $p$ is even and $q$ is odd, most points will pair up like in the previous case, but there will be a leftover collection of points $\sqrt{\frac{b}{a}} (2m-1)$ which land on even integers. Fortunately, the tangent is zero on these points, so the sum will again converge like before.

So summarily the sum converges only when $\sqrt{\frac{b}{a}}$ is a rational number with either the numerator or denominator even. When it does converge, you should be able to find an explicit formula for the sum by adding up a finite number of sums of the form $\sum \frac{1}{a+n^2}$ (arising from the pairing of points mentioned above), which can each be evaluated analytically.

(This is quite sketchy, but it doesn't apply to your particular value anyway, so I'm hoping I can get away with it...)

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Let

$$q=\sqrt{\dfrac{b}{a}},$$ then attack via digamma-function leads to $$\begin{align} &S = \dfrac1a\sum\limits_{n=1}^\infty\sum\limits_{m=1}^\infty\dfrac1{(2n-1)^2-q^2(2m-1)^2} \\[4pt] &= \dfrac1a\sum\limits_{n=1}^\infty\sum\limits_{m=1}^\infty\dfrac1{2(2n-1)}\left(\dfrac1{2n-1-q(2m-1)}+\dfrac1{2n-1+q(2m-1)}\right)\\[4pt] &= \dfrac1{4aq}\sum\limits_{n=1}^\infty\dfrac1{2n-1}\sum\limits_{m=1}^\infty \left(-\dfrac1{m-\frac{2n-1+q}{2q}}+\dfrac1{m+\frac{2n-1-q}{2q}}\right)\\[4pt] &= \dfrac1{4aq}\sum\limits_{n=1}^\infty\dfrac1{2n-1}\left(\psi\left(1-\frac{2n-1+q}{2q}\right)-\psi\left(1+\frac{2n-1-q}{2q}\right)\right)\\[4pt] &= \dfrac1{4aq}\sum\limits_{n=1}^\infty\dfrac1{2n-1}\left(\psi\left(\frac{q-2n+1}{2q}\right)-\psi\left(\frac{q+2n-1}{2q}\right)\right)\\[4pt] &= \dfrac1{4aq}\sum\limits_{n=1}^\infty\dfrac1{2n-1}\cdot\pi\cot\left(\pi\frac{q-2n+1}{2q}\right)\\[4pt] &= \dfrac\pi{4\sqrt{ab}}\sum\limits_{n=1}^\infty\dfrac1{2n-1}\tan\left(\frac\pi 2 \sqrt{\frac ab}(2n-1)\right)\\[4pt] &= \dfrac\pi{4\sqrt{ab}}\sum\limits_{m=1}^\infty\dfrac1{2m-1}\tan\left(\frac\pi 2 \sqrt{\color{red}{\frac ab}}(2m-1)\right). \end{align}$$

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  • $\begingroup$ (+1) $$ \begin{align}S&=\color{red}{+}\frac{\pi}{4\sqrt{ab}}\sum_{\color{red}{m=1}}^\infty\frac{\tan\left(\frac{\pi}{2}\sqrt{\color{red}{\frac{b}{a}}}\left(2m-1\right)\right)}{2m-1}\\&=\color{red}{-}\frac{\pi}{4\sqrt{ab}}\sum_{\color{red}{n=1}}^\infty\frac{\tan\left(\frac{\pi}{2}\sqrt{\color{red}{\frac{a}{b}}}\left(2n-1\right)\right)}{2n-1}\end{align} $$ $\endgroup$ – Hazem Orabi Jan 29 '19 at 13:33
  • $\begingroup$ @HazemOrabi I've elaborated my reasons in the answer. What are yours? $\endgroup$ – Yuri Negometyanov Jan 29 '19 at 17:14

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