How fast the solution of $\frac{e^x}{e^x-1}-\frac{1}{x}=\frac{1}{n}$ goes to $-\infty$ as $n\to\infty$? I have an equation, for each $n\in\mathbb{N}$, $n\geq 3$:
$$
f(x):=\frac{e^x}{e^x-1}-\frac{1}{x} = \frac{1}{n}.
$$
which has a unique solution which tends to $-\infty$ as $n\to\infty$, since the left-hand side is continuous and monotone increasing with the limit $\lim_{x\to\pm\infty} f(x)= 1$ and $0$.
Since the solution cannot be written in a closed form, I would like to know how fast the solution goes to $-\infty$ comparing to $n$. In other words, how can I expressed the solution in terms of $n$, say $\phi(n)$, so that $f(\phi(n))\sim\frac{1}{n}$ as $n\to\infty$?
 A: For large $n$ if we take $\phi(n) = -n$ then $$\left|f(-n) - \frac{1}{n}\right| =\left|\frac{1}{1-e^n}\right|<2e^{-n}.$$
A: Since the solution cannot be written in a closed form...
$$\frac{e^x}{e^x-1}-\frac{1}{x} = \frac{1}{n}\quad \implies \quad e^{-x}=\frac {n+(1-n)x}{x+n}$$ has an explicit solution in terms  of the generalized Lambert function (have a look at equation $(4)$).
If you consider the reciprocal for of the equation, that is to say
$$\frac{\left(e^x-1\right) x}{e^x (x-1)+1}=n$$ the solution is close to $x_0=-n$
The first iterate of Newton method is
$$x_1=-n+\frac{\left(e^n-1\right) n^2}{1+e^{2 n}-e^n \left(n^2+2\right)}$$ which seems to be decent.
For illustration purposes, let $n=2^k$ and compare
$$\left(
\begin{array}{ccc}
 n & x_1 & \text{solution} \\
 4 & -3.5710413473964268697 &  -3.5935119694474260822 \\
 8 & -7.9780516483210477777 &  -7.9781077369770703258 \\
 16 & -15.999971190162046898 &  -15.999971190213911766 \\
 32 & -31.999999999987031894 &  -31.999999999987031894 \\
\end{array}
\right)$$
Neglecting the "small" numbers
$$x_1 \sim -n + n^2\,e^{-n}$$
