Is it true that


$$f(x)=\lim\_{n\to\infty} 2 \sum \_{k=0}^n \left((k-1) \text{Li}\_k\left(\frac{f(x)}{n^2}\right)-x \text{Li}\_{k-1}\left(\frac{f(x)}{n^2}\right)\right)?$$


Here, $f(x)$ is an arbitrary function that I tested. I found this by chance, but numerically it looks OK (tried 5000 terms with $\exp$, $\cosh$, $\sinh$ and $\sin$). Is there any justification?



This equality numerically looks well. Is there any justification?
$\newcommand\Li{\text{Li}}$This follows immediately because uniformly over all $k\ge-1$ we have $\Li\_k(z)\sim z$ as $z\to0$, where $\Li\_k$ is the [polylogarithm function](https://en.wikipedia.org/wiki/Polylogarithm).




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**Details:** For $k\ge-1$ and $|z|\downarrow0$,
$$\Big|\frac{\Li\_k(z)}z-1\Big|=\Big|\frac z{2^k}+\frac{z^2}{3^k}+\cdots\Big|\le|z|\sum\_{j=0}^\infty(j+2)|z|^j\to0$$
So, for $y:=f(x)$ and $n\to\infty$,
$$\Big|\sum\_{k=0}^n x\Li\_{k-1}(y/n^2)\Big|=O(|xy|(n+1)/n^2)\to0$$
and
$$2\sum\_{k=0}^n (k-1)\Li\_k(y/n^2)
=(1+o(1))\frac y{n^2}\,\sum\_{k=0}^n 2(k-1)\to y.$$


