Non-trivial zero to an infinite series In my free time, I was messing around with some maths, which eventually led me to the question of finding a non-trivial zero of the following series:
$$\sum_{n=1}^{\infty} \frac{nx^n}{1-x^n} $$
Does anybody maybe know a closed form (probably not)? Or does someone maybe know whether it has been studied before and has a name? If someone can help, that would be appreciated! (The zero I am looking for is around -0.41)
 A: While I do not have a closed-form solution, I was able to solve it numerically using the Newton-Raphson method with the help of a Java program I coded. The answer is approximately 
                                      x = -0.4112484443599058

If this value is plugged into the function provided, the $y$ value is within $\pm0.0000001$ of $0$. Sorry I could not provide a closed-form solution. The code for the Java program is attached if you would like to check it. 
Source Code
I hope this helps!!! :)
A: We can rewrite the sum as $$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\left(nx^{mn}\right)$$
Switching the order of summation and evaluating, we get $$\sum_{m=1}^{\infty}\frac{x^{m}}{\left(x^{m}-1\right)^{2}}$$
Now WolframAlpha does not give a closed form for this, but it does give partial sums. Using that, the above sum is $$\lim_{n \to \infty} \frac{\psi_x^{(1)}(1)-\psi_x^{(1)}(n)}{\ln^2(x)}$$
Setting this equal to $0$ and simplifying, we get $$\psi_x^{(1)}(1) = \lim_{n \to \infty} \psi^{(1)}_x(n)$$
I think that $\psi_x(n)$ approaches a constant as $n$ approaches $\infty$, which makes the  RHS equal to $0$. Therefore, you want to solve $$\psi_x^{(1)}(1) = 0$$
Searching up the first few digits on the OEIS, I found A143441. This gives a value of $$-0.41124847917795477344...$$ which when plugged into the original equation provides an error of $-1.14 \cdot 10^{-17}$.
