Showing the root of $\sum_{j=1}^n \binom{n}{j} \frac{j x^{j-1}}{(1-x^j)^2}$ approaches $0$ as $n \to \infty$. 
Let $n > 1$, and let $\varepsilon_n$ be the root of $g_n(x) = \sum_{j=1}^n \binom{n}{j} \frac{j x^{j-1}}{(1-x^j)^2}$ that is in the interval $(-1, 1)$. Why is $\lim_{n \to \infty} \varepsilon_n = 0$?

Here, let's agree to interpret $x^{j-1}$ as $1$ if $j=1$ (even if $x=0$).
This would answer one of the questions I asked in this other question. Specifically, this would show that the $p_n$ in my other question does approach $1/2$; (see Claude Leibovici's answer).

Edit:
The accepted answer gives a proof for this with one pertinent detail missing, which is filled in here in this other question.

An alternative representation of $g_n$, and a simple estimate, suffice.
For $|x|<1$, put $f_n(x)=\sum_{j=1}^n\binom{n}{j}\frac1{1-x^j}$, so that $g_n(x)=f_n'(x)$. Since $$f_n(x)=\sum_{j=1}^n\binom{n}{j}\sum_{k=0}^\infty x^{jk}=\sum_{k=0}^\infty\big((1+x^k)^n-1\big),$$ we get $g_n(x)=n\sum_{k=1}^\infty kx^{k-1}(1+x^k)^{n-1}$. Now put, for convenience, $$G_n(x)=\frac{g_{n+1}(-x)}{n+1}=(1-x)^n-2x(1+x^2)^n+3x^2(1-x^3)^n-4x^3(1+x^4)^n+\dots$$ and $\xi_n=-\varepsilon_{n+1}$ (so that $0<\xi_n<1$ and $G_n(\xi_n)=0$; we're showing $\xi_n\to 0$ as $n\to\infty$).
Clearly $G_{n+1}(x)<G_n(x)$ for $0<x<1$; hence $\xi_{n+1}<\xi_n$. Now the estimate is $$G_n(x)<(1-x)^n-2x+3x^2-4x^3+\dots=(1-x)^n+(1+x)^{-2}-1,$$ and $\xi_n\to 0$ is easy to see: otherwise there is $\xi>0$ such that $\xi_n>\xi$ for all $n$, then $G_n(\xi)>0$; but, using the estimate above, and taking $n\to\infty$, we get $(1+\xi)^{-2}-1\geqslant 0$, which is absurd.
In fact it can be shown that $\xi_n=O\big(\frac{\log n}{n}\big)$.
