Showing the root of $\sum_{j=1}^n \binom{n}{j} \frac{j x^{j-1}}{(1-x^j)^2}$ approaches $0$ as $n \to \infty$.

Question

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).

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

Answer 1

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 $\xi_n=O\big(\frac{\log n}{n}\big)$.

Answer 2

In my previous answer, I missed something important : we need to solve palindromic polynomials. All of them are of even degree and only show positive coefficients.

$$\left( \begin{array}{cc} n & \text{coefficients} \\ 2 & 1,3,1\\ 3 & 1,6,13,18,13,6,1 \\ 4 & 1,7,19,40,54,67,54,40,19,7,1 \\ \end{array} \right)$$

Starting from $n=2$, the degrees of these polynomials are $$\{2,6,10,18,22,34,42,54,62,82,90,114,126,142,158,190,202,238,254\}$$

In the real domain, they show only two negative roots the product of which being strictly equal to $1$. The small one $x_2$ is the one of interest and the other one $x_1$ tends to $-\infty$.

$$\left( \begin{array}{cc} n & x_1 & x_2 \\ 2 & -2.61803& -0.381966 \\ 3 & -3.48211& -0.287182 \\ 4 & -4.19294& -0.238496 \\ 5 & -4.82329& -0.207327 \\ 6 & -5.40199& -0.185117 \\ 7 & -5.94387& -0.168241 \\ 8 & -6.45778& -0.154852 \\ 9 & -6.94948& -0.143896 \\ 10 & -7.42297& -0.134717 \\ \end{array} \right)$$

Working with $x=x_1$ and using for large values of $x$ $$ \frac{ x^{j-1}}{(1-x^j)^2}=\sum_{k=1}^\infty \frac k{x^{kj+1}}$$ Using the first and second terms, we end with the equation $$x^n (x+1)^{n-1}+2 \left(x^2+1\right)^{n-1}=0$$ the solution of which being the inverse of $$n=\frac{\log \left(-\frac{x^2+1}{2 (x+1)}\right)}{\log \left(\frac{x^2+1}{x (x+1)}\right)}$$

Expanding the rhs

$$n=x \log \left(-\frac{2}{x}\right)+\frac{1}{2} \left(3 \log \left(-\frac{2}{x}\right)+2\right)+\frac{23 \log \left(-\frac{2}{x}\right)}{12 x}+\cdots$$

Limited to $O\left(\frac{\log \left(-\frac{2}{x}\right)}{x}\right)$, there is a solution in terms of the generalized Lambert function since letting $x=-2e^t$ the equation write $$e^{-t}=\frac43\frac{ t}{t+\frac{2(n-1)}3}$$

Using only the first term gives as an estimate of this approximate equation $$x_1^{(0)}=-\frac{n}{W\left(\frac{n}{2}\right)}\qquad \implies\qquad \color{red}{\large x_2^{(0)}=-\frac{W\left(\frac{n}{2}\right)}n}$$ where $W(.)$ is Lambert function.

For $n=10$, this gives $x_2^{(0)}=-0.132672$ the exact solution being $x=−0.134717$.

For $10 \leq n \leq 20$, the degree of correlation between the estimate and the solution is $0.999955$.

Numerical aspects

Noting $x_1=x_1^{(0)}$, even if it does not mean much

$$g_n(x_1) = \sum_{j=1}^n \binom{n}{j} j\,\frac{x_1^{j-1}}{(1-x_1^j)^2}$$ is numerically small.

$$\forall n \geq 12 \qquad \qquad -10^{-3} < g_n(x_1) <10^{-3}$$ and it is even asymptotic to $0^-$.

Solving $g_n(x)=0$ without any transform, Newton method converges very fast. For example, for $n=123$, the successive iterates are $$\left( \begin{array}{cc} k & x_k \\ 0 & -40.7915765085 \\ 1 & -41.2964036355 \\ 2 & -41.3088991630 \\ 3 & -41.3089065467 \\ \end{array} \right)$$