# This equality numerically looks well. Is there any justification?

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?

• The use of $f(x)$ here is very strange... Jun 24, 2021 at 16:15
• Could you introduce your notation and use quantifiers? $f$ is not introduced, $x$ is not introduced.
– YCor
Jun 24, 2021 at 17:05
• Could it be a Riemann sum? The parenthesis can be rewritten by using $\Li_{k-1}(f(x)/n^2)=(f(x)/n^2) \frac{d \Li_k(f(x) /n^2)}{d f(x) /n^2}$ Jun 24, 2021 at 17:08
• As mentioned in a comment that was for some reason deleted, can't you just replace $f(x)$ by $x$? Jun 24, 2021 at 17:09
• Since $f$ and $x$ are numbers that can be varied independently, the question really decomposes into two separate questions: The second term on the r.h.s. must sum to zero, the first term on the r.h.s. must sum to $f$. Jun 24, 2021 at 18:09

$$\newcommand\Li{\text{Li}}$$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.
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 $$2\sum_{k=0}^n (k-1)\Li_k(y/n^2) =(1+o(1))\frac y{n^2}\,\sum_{k=0}^n 2(k-1)\to y.$$