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.

## Comments

Sort by:

TopNewestSimilarly \[\mathrm{Li}_{-m}(z)=\frac{\left(-1\right)^m}{1-z}\left[\sum _{k=0}^{m-1}\left(-1\right)^k\binom{m}{k}\mathrm{Li}_{-k}(z)\right]\] – Julian Poon · 1 year, 2 months ago

Log in to reply

– Jake Lai · 1 year, 2 months ago

Nice!Log in to reply

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. – Ishan Singh · 8 months, 3 weeks ago

Log in to reply