Please allow me to resort once again to the expertise of the MathOverflow community :

During research I encoutered the following infinite series :

$$\sum_{n=-\infty}^{+\infty} \frac{u^{2n}}{1+\rho^{2n+1}},$$ where $0<\rho<1$,$-1<u<1$ and $\rho<|u|$.

Is this series related to some well-known special functions, for instance the elliptic functions or theta functions? I have reasons to believe that it should be.

Thank you,

Best regards, Malik

Note Unless I made a mistake, the above sum can be rewritten as $$\sum_{m=0}^{\infty} (-1)^m \frac{\rho^m}{1-u^2\rho^{2m}} + \sum_{m=0}^{\infty} (-1)^m \frac{u^{-2}\rho^{m+1}}{1-u^{-2}\rho^{2m+2}},$$ which suspicously is quite similar to the infinite series appearing in the formulas for the expansion of the Weierstrass elliptic functions in term of the elliptic nome (see e.g. Akhiezer's Elements of the theory of elliptic functions, p.204).

EDIT (In response to Alexandre Eremenko's comment)

The series is convergent. Indeed, the $n<0$ part can be written

$$\sum_{n=1}^{\infty} \frac{u^{-2n}}{1+\rho^{-2n+1}} = \sum_{n=1}^{\infty} \frac{\rho^{2n}u^{-2n}}{\rho^{2n}+\rho}$$ which is clearly convergent since $\rho<|u|$.

  • $\begingroup$ Obvious comment: have you tried feeding it to Mathematica or Maple or some similar CAS? $\endgroup$ – Nate Eldredge Nov 5 '14 at 20:23
  • $\begingroup$ @NateEldredge I tried with Maple and didn't get anything, but it might just mean I don't know how to use Maple properly! $\endgroup$ – Malik Younsi Nov 5 '14 at 20:58

This is a special case of Ramanujan's ${}_1\psi_1$ summation, which sums the series $$\sum_{-\infty}^\infty\frac{(a;q)_n}{(b;q)_n}\,x^n,$$ where $(a;q)_n=(1-a)(1-aq)\dotsm (1-a q^{n-1})$. If we let $b=aq$, the series simplifies to $$\sum_{n=-\infty}^\infty\frac{1-a}{1-aq^n}\,x^n,$$ which is equivalent to yours. The answer can be written $$\sum_{n=-\infty}^\infty\frac{x^n}{1-aq^{2n}}=\frac{(q^2;q^2)_\infty^2\theta(ax;q^2)}{\theta(a;q^2)\theta(x;q^2)},$$ which holds for $|q|<|x|<1$, where $\theta(x;q)=(x;q)_\infty(q/x;q)_\infty$. If you rewrite this theta in terms of Jacobi's $\vartheta_1$ you get Edgar's answer.

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I like these... $$ \sum _{n=-\infty }^{\infty }{\frac {{x}^{n}}{1-a{q}^{2\,n}}} ={\frac {- \frac{1}{2}\,i\vartheta_1 \left( \frac{1}{2}\,i\ln \left( ax \right) ,q \right) \vartheta_2 \left( 0,q \right) \vartheta_3 \left( 0,q \right) \vartheta_4 \left( 0,q \right) }{\vartheta_1 \left( \frac{1}{2}\,i\ln \left( x \right) ,q \right) \vartheta_1 \left( \frac{1}{2}\,i\ln \left( a \right) ,q \right) }} $$

$$ \sum _{n=-\infty }^{\infty } \frac{1}{ {\alpha}^{n}-a{\beta}^{n} } =\frac{-\frac{1}{2}\,i\vartheta_1 \left( \frac{1}{2}\,i\ln \left( {\frac { \alpha}{a}} \right) ,\sqrt {{\frac {\beta}{\alpha}}} \right) \vartheta_2 \left( 0,\sqrt {{\frac {\beta}{\alpha}}} \right) \vartheta_3 \left( 0,\sqrt {{\frac {\beta}{\alpha}}} \right) \vartheta_4 \left( 0,\sqrt {{\frac {\beta}{\alpha}}} \right) }{ \vartheta_1 \left( \frac{1}{2}\,i\ln \left( \alpha \right) , \sqrt {{\frac {\beta}{\alpha}}} \right) \vartheta_1 \left( \frac{1}{2}\,i\ln \left( a \right) ,\sqrt {{\frac {\beta }{\alpha}}} \right) } $$

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  • 2
    $\begingroup$ Whoa, where does that come from? :-) $\endgroup$ – Malik Younsi Nov 5 '14 at 21:22
  • $\begingroup$ Not from Mathematica (I tried). $\endgroup$ – Igor Rivin Nov 5 '14 at 22:11
  • $\begingroup$ The formula looks correct, but I have no idea how to obtain it... $\endgroup$ – Malik Younsi Nov 5 '14 at 22:30
  • 1
    $\begingroup$ What are $\alpha$ and $\beta$? $\endgroup$ – Robert Israel Nov 5 '14 at 23:07
  • $\begingroup$ $0 < |q|^2 < |x| < 1$ and $0 < |\beta| < 1 < |\alpha|$. $\endgroup$ – Gerald Edgar Nov 8 '14 at 14:12

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