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Is it possible to obtain the sum of the infinite series: $$\sum_{n=1}^\infty \frac{c^n}{1-q^n}$$ where $0<c<1, 0<q<1$.

more information:

I want to prove that the imaginary part of the following complex function $f(z)$ is constant on the circle $C=\{z|z=b+ae^{i\theta},0<a<b,0\le\theta<2\pi\}$: $$f(z)=-iz+2iza^2\left\{\frac{1}{z^2-b^2}+\sum_{n=1}^\infty \frac{a^{2n}}{\prod_{r=0}^{n-1}(b+x_r)^2\cdot(z^2-x_n^2)}\right\}$$ where $x_0=b,x_n=b-{a^2}/(b+x_{n-1})$ , so that $f(z)$ descripes a potential flow around the circle.

I transformed $f(z)$ into a simpler form: $$f(z)=-iz+2ik\sum_{m=1}^\infty \left[ \left(\frac{z+k}{z-k}q\right)^m - \left( \frac{z-k}{z+k}q\right)^m\right]\frac{1}{1-q^m}$$ where $k=\sqrt{b^2-a^2},q=(b-k)/(b+k)$,which contains infinite seires of the form $\sum \frac{c^n}{1-q^n}$(here $c$ is actually a complex number). I tried to find the summation of the seires but failed.

Thanks for user10354138's answer, I give up finding the sum of this series and change $f(z)$ into another form and finally succeed in proving the constantness of the imaginary part of $f(z)$ on the circle.

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    $\begingroup$ Depends on what you mean. Clearly the sum is finite, but there are no elementary closed form expression for it in terms of $c,q$. $\endgroup$ – user10354138 May 22 at 12:54
  • $\begingroup$ Interestingly (but probably uselessly) it can also be written $$\sum_{n=1}^\infty \frac{c^n}{1-q^n}=\frac{1}{1-c}+\sum_{n=1}^{\infty}\left(\sum_{d\mid n}c^d\right)q^n.$$ $\endgroup$ – Servaes May 22 at 14:12
  • $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. $\endgroup$ – dantopa May 22 at 22:09
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With CAS help:

$$\sum _{n=1}^{\infty } \frac{c^n}{1-q^n}=\sum _{m=0}^{\infty } \left(\sum _{n=1}^{\infty } c^n \left(q^n\right)^m\right)=\sum _{m=0}^{\infty } -\frac{c q^m}{-1+c q^m}=\frac{\ln (1-q)}{\ln (q)}+\frac{\psi _q\left(\frac{\ln (c)}{\ln (q)}\right)}{\ln (q)}$$

where: ${\psi _q\left(\frac{\ln (c)}{\ln (q)}\right)}$ is q-digamma function.

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