Non-trivial zero to an infinite series

Question

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)

Answer 1

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 $-1.14 \cdot 10^{-17}$.

Answer 2

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!!! :)

Answer 3

To have approximations of the solution of equation $$\psi_x^{(1)}(1) = 0$$ we could build around $x=0$ the $[1,n]$ Padé approximant of $$\frac{\psi_x^{(1)}(1)}{x \, \log^2(x)}$$ this will generate for the approximate solution the sequence $$-\frac{3}{5},-\frac{1}{2},-\frac{2}{5},-\frac{25}{64},-\frac{2}{5},-\frac{16 }{39},-\frac{39}{94},-\frac{94}{227},-\frac{454}{1103},-\frac{1103}{2688},-\frac {2688}{6547},$$ $$-\frac{6547}{15922},-\frac{7961}{19348},-\frac{96740}{235153},-\frac {470306}{1143585},-\frac{228717}{556214},-\frac{278107}{676299},-\frac{225433}{ 548170},$$ $$-\frac{93972}{228499},-\frac{228499}{555614},-\frac{9723245}{23643206},-\frac{23643206}{57491577},-\frac{57491577}{139798147},\cdots,$$ $$-\frac{2933110593712303331274025997834035269997622991978006995444}{7132210189750251 897714249395557962621504069786938318973765},\cdots$$