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?

  • 3
    $\begingroup$ The use of $f(x)$ here is very strange... $\endgroup$ Jun 24, 2021 at 16:15
  • 1
    $\begingroup$ Could you introduce your notation and use quantifiers? $f$ is not introduced, $x$ is not introduced. $\endgroup$
    – YCor
    Jun 24, 2021 at 17:05
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    $\begingroup$ Could it be a Riemann sum? The parenthesis can be rewritten by using $\Li_{k-1}(f(x)/n^2)=(f(x)/n^2) \frac{d \Li_k(f(x) /n^2)}{d f(x) /n^2}$ $\endgroup$
    – Archie
    Jun 24, 2021 at 17:08
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    $\begingroup$ As mentioned in a comment that was for some reason deleted, can't you just replace $f(x)$ by $x$? $\endgroup$ Jun 24, 2021 at 17:09
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    $\begingroup$ 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 $f$. $\endgroup$ Jun 24, 2021 at 18:09

1 Answer 1


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


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