# Is it possible to obtain the sum of this infinite series?

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

I want to prove that the imaginary part of the following complex function $$f(z)$$ is constant on the circle $$C=\{z|z=b+ae^{i\theta},0: $$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.

• Depends on what you mean. Clearly the sum is finite, but there are no elementary closed form expression for it in terms of $c,q$. – user10354138 May 22 at 12:54
• Interestingly (but probably uselessly) it can also be written $$\sum_{n=1}^\infty \frac{c^n}{1-q^n}=\frac{1}{1-c}+\sum_{n=1}^{\infty}\left(\sum_{d\mid n}c^d\right)q^n.$$ – Servaes May 22 at 14:12
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$$\sum _{n=1}^{\infty } \frac{c^n}{1-q^n}=\sum _{m=0}^{\infty } \left(\sum _{n=1}^{\infty } c^n \left(q^n\right)^m\right)=\sum _{m=0}^{\infty } -\frac{c q^m}{-1+c q^m}=\frac{\ln (1-q)}{\ln (q)}+\frac{\psi _q\left(\frac{\ln (c)}{\ln (q)}\right)}{\ln (q)}$$
where: $${\psi _q\left(\frac{\ln (c)}{\ln (q)}\right)}$$ is q-digamma function.