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# Polylogarithm recursion identity

I recently discovered a recursive identity for $$\mathrm{Li}_{-m}(z)$$ for $$m$$ a positive integer. Hopefully this will be of some use when tackling problems seeking for the value of some polylogarithmic sum.

$\mathrm{Li}_{-m}(z) = \frac{z}{1-z} \left[ 1+\sum_{k=0}^{m-1} \binom{m}{k} \mathrm{Li}_{-k}(z) \right]$

Proof:

Consider $$\displaystyle \sum_{n=1}^{\infty} z^n(n+1)^m$$. We have

\begin{align} \sum_{n=1}^{\infty} z^n(n+1)^m &= \sum_{n=1}^{\infty} z^n \sum_{k=0}^m \binom{m}{k} n^k \\ &= \sum_{n=1}^{\infty} z^nn^m + \sum_{k=0}^{m-1} \binom{m}{k} \sum_{n=1}^{\infty} z^nn^k \\ &= \mathrm{Li}_{-m}(z) + \sum_{k=0}^{m-1} \binom{m}{k} \mathrm{Li}_{-k}(z) \end{align}

We also have

\begin{align} \sum_{n=1}^{\infty} z^n(n+1)^m &= \frac{1}{z} \sum_{n=2}^{\infty} z^nn^m \\ &= \frac{1}{z} \mathrm{Li}_{-m}(z)-1 \end{align}

Equating the two gives

$\mathrm{Li}_{-m}(z) + \sum_{k=0}^{m-1} \binom{m}{k} \mathrm{Li}_{-k}(z) = \frac{1}{z} \mathrm{Li}_{-m}(z)-1$

When rearranged, we obtain

$\mathrm{Li}_{-m}(z) = \frac{z}{1-z} \left[ 1+\sum_{k=0}^{m-1} \binom{m}{k} \mathrm{Li}_{-k}(z) \right]$

as required.

Note by Jake Lai
10 months, 3 weeks ago

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Similarly $\mathrm{Li}_{-m}(z)=\frac{\left(-1\right)^m}{1-z}\left[\sum _{k=0}^{m-1}\left(-1\right)^k\binom{m}{k}\mathrm{Li}_{-k}(z)\right]$ · 10 months, 3 weeks ago

Nice! · 10 months, 3 weeks ago

A point to note is that $$\operatorname{Li}_{-m} (z)$$ , when $$m$$ is a positive integer, already has a closed form in terms of rational polynomial functions. · 4 months, 3 weeks ago