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I have the following expression

$$ \sum_{n=-\infty}^{+\infty} \frac{1}{(1-q^{2n-1})(1-q^{m-2n})}$$ where $m$ is an odd positive integer and $|q|<1$.

Given this, the sum should converge to a finite answer, which I would like to compute.

As $|q|<1$ I can expand the reciprocal factors in power series depending on whether $n\in (-\infty,0)$, $n\in (1,\frac{m-1}{2})$ and $n\in(\frac{m+1}{2},+\infty)$. Then by summing over the $n$ index I am left with the following expression:

$$\sum_{a,b=0}^{+\infty}\frac{q^{a+mb}-q^{ma+b}}{q^{2b}-q^{2a}} - \frac{q^{1+a+mb}+q^{1+ma+b}}{1-q^{2(a+b+1)}}$$

What I would like to do now is to compute the sums over $a,b$ but I can't seem to figure out how to do this.

Any help would be greatly appreciated.

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  • $\begingroup$ Many ${\space}$ Thanks. $\endgroup$ – Hazem Orabi Sep 26 '18 at 6:30
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$ \underline{\text{For}\space\left(m\ne1\right)\space\colon} $

$$ \begin{align} \color{red}{S(m,q)} &=\sum_{-\infty}^{+\infty}\frac{1}{\left(1-q^{2n-1}\right)\left(1-q^{m-2n}\right)} \\ &=\sum_{-\infty}^{+\infty}\frac{-q^{-2n-m}}{\left(q^{-2n}-q^{-1}\right)\left(q^{-2n}-q^{-m}\right)} \\ &=\frac{-q^{-m}}{q^{-1}-q^{-m}}\sum_{-\infty}^{+\infty}\left[\frac{q^{-2n}}{q^{-2n}-q^{-1}}-\frac{q^{-2n}}{q^{-2n}-q^{-m}}\right] \\ &=\frac{1}{1-q^{m-1}}\sum_{-\infty}^{+\infty}\left[\frac{1}{1-q^{2n-1}}-\frac{1}{1-q^{2n-m}}\right] \\ &=\frac{\color{blue}{S^{-}(m,q)}\,+\,\color{magenta}{S^{+}(m,q)}}{1-q^{m-1}} =\color{red}{\frac{(m-1)/2}{1-q^{m-1}}} \\[6mm] \color{blue}{S^{-}(m,q)} &=\sum_{n=-1}^{-\infty}\left[\frac{1}{1-q^{2n-1}}-\frac{1}{1-q^{2n-m}}\right] \\ &=\sum_{n=1}^{\infty}\left[\frac{1}{1-q^{-2n-1}}-\frac{1}{1-q^{-2n-m}}\right] \\ &=\sum_{n=1}^{\infty}\frac{1}{1-q^{-2n-1}}-\sum_{n=1}^{\infty}\frac{1}{1-q^{-2n-m}} \\ &=\sum_{n=1}^{\infty}\frac{1}{1-q^{-2n-1}}-\sum_{n=+\frac{m+1}{2}}^{\infty+\frac{m+1}{2}}\frac{1}{1-q^{-2n-1}} \\ &=\sum_{n=1}^{\frac{m-1}{2}}\frac{1}{1-q^{-2n-1}}-\lim_{l\to\infty}\sum_{n=l+1}^{l+\frac{m+1}{2}}\frac{1}{1-q^{-2n-1}} \\ &=\color{blue}{\sum_{n=1}^{\frac{m-1}{2}}\frac{1}{1-q^{-2n-1}}\quad\qquad\left\{\lim_{x\to\infty}\frac{1}{1-q^{-x}}=0\space\colon|q|\lt1\right\}} \\[6mm] \color{magenta}{S^{+}(m,q)} &=\sum_{n=0}^{\infty}\left[\frac{1}{1-q^{2n-1}}-\frac{1}{1-q^{2n-m}}\right] \\ &=\sum_{n=0}^{\infty}\frac{1}{1-q^{2n-1}}-\sum_{n=0}^{\infty}\frac{1}{1-q^{2n-m}} \\ &=\sum_{n=0}^{\infty}\frac{1}{1-q^{2n-1}}-\sum_{n=-\frac{m-1}{2}}^{\infty-\frac{m-1}{2}}\frac{1}{1-q^{2n-1}} \\ &=\lim_{l\to\infty}\sum_{l-\frac{m-3}{2}}^{l}\frac{1}{1-q^{2n-1}}-\sum_{-\frac{m-1}{2}}^{-1}\frac{1}{1-q^{2n-1}} \\ &=\color{magenta}{\frac{m-1}{2}-\sum_{n=1}^{\frac{m-1}{2}}\frac{1}{1-q^{-2n-1}}\quad\left\{\lim_{x\to\infty}\frac{1}{1-q^{x}}=1\space\colon|q|\lt1\right\}} \end{align} $$


$ \underline{\text{For}\space\left(m=1\right)\space\colon} $
It is interesting how the summation takes a totally different path for $m=1$. Using the above result, we have: $$ \lim_{m\to1}\frac{(m-1)/2}{1-q^{m-1}} =\frac{-1}{2\log(q)}\quad\ne\,S(1,q) $$ This can be seen from the main series: $$ \begin{align} \color{red}{S(1,q)}&=\sum_{-\infty}^{+\infty}\frac{1}{\left(1-q^{2n-1}\right)\left(1-q^{1-2n}\right)} =\sum_{-\infty}^{+\infty}\frac{-q^{-2n+1}}{\left(1-q^{-2n+1}\right)^2} \\ &=\sum_{-\infty}^{0}\frac{-q^{-2n+1}}{\left(1-q^{-2n+1}\right)^2}+\sum_{1}^{\infty}\frac{-q^{-2n+1}}{\left(1-q^{-2n+1}\right)^2} \\ &=\sum_{n=0}^{\infty}\frac{-q^{-2n-1}}{\left(1-q^{-2n-1}\right)^2}+\sum_{n=0}^{\infty}\frac{-q^{-2n-1}}{\left(1-q^{-2n-1}\right)^2} \\ &=-2\sum_{n=0}^{\infty}\frac{\left(q^2\right)^{-n-\frac12}}{\left(1-\left(q^2\right)^{-n-\frac12}\right)^2} =\color{red}{-\frac{2\,\psi^{(1)}_{q^2}(\frac12)}{\log^2\left(q^2\right)}} \end{align} $$ And the sum is related to q-Polygamma function, which is the derivative of q-Digamma function: $$ \begin{align} \psi_{q}(z) &=-\log(1-q)+\log(q)\sum_{n=0}^{\infty}\frac{q^{n+z}}{1-q^{n+z}} \\ &=-\log(1-q)-\log(q)\sum_{n=0}^{\infty}\frac{1}{1-q^{-n-z}} \\ \psi^{(m)}_{q}(z)&=\frac{d^m}{dz^m}\left[\psi_{q}(z)\right] \implies \psi^{(1)}_{q}(z)=\log^2(q)\sum_{n=0}^{\infty}\frac{q^{-n-z}}{\left(1-q^{-n-z}\right)^2} \end{align} $$

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  • $\begingroup$ do not get $S(m,q)=\sum_{-\infty}^{+\infty}\frac{1}{\left(1-q^{2n-1}\right)\left(1-q^{m-2n}\right)} =\sum_{-\infty}^{+\infty}\frac{-q^{-2n-m}}{\left(q^{-2n}-q^{-1}\right)\left(q^{-2n}-q^{-m}\right)}$ $\endgroup$ – G Cab Sep 14 '18 at 17:24
  • $\begingroup$ @GCab : $$ \frac{1}{\left(1-q^{2n-1}\right)\left(1-q^{m-2n}\right)} \\ =\color{red}{\frac{q^{-2n}}{q^{-2n}}}\,\times\,\frac{1}{\color{red}{\left(1-q^{2n-1}\right)}\color{blue}{\left(1-q^{m-2n}\right)}}\times\color{blue}{\frac{q^{-m}}{q^{-m}}} \\ =\frac{q^{-2n-m}}{\left(q^{-2n}-q^{-1}\right)\left(q^{-m}-q^{-2n}\right)} $$ $\endgroup$ – Hazem Orabi Sep 15 '18 at 0:36
  • $\begingroup$ oh yes, sorry: my (temporary..) blindness. $\endgroup$ – G Cab Sep 15 '18 at 13:32

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