Polylogarithm recursion identity

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

Li−m(z)=z1−z[1+∑k=0m−1(mk)Li−k(z)]\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 ∑n=1∞zn(n+1)m\displaystyle \sum_{n=1}^{\infty} z^n(n+1)^m. We have

∑n=1∞zn(n+1)m=∑n=1∞zn∑k=0m(mk)nk=∑n=1∞znnm+∑k=0m−1(mk)∑n=1∞znnk=Li−m(z)+∑k=0m−1(mk)Li−k(z)\begin{aligned} \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{aligned}

We also have

∑n=1∞zn(n+1)m=1z∑n=2∞znnm=1zLi−m(z)−1\begin{aligned} \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{aligned}

Equating the two gives

Li−m(z)+∑k=0m−1(mk)Li−k(z)=1zLi−m(z)−1\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

Li−m(z)=z1−z[1+∑k=0m−1(mk)Li−k(z)]\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.

#Calculus

Note by Jake Lai
5 years, 7 months ago

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Comments

Similarly Li−m(z)=(−1)m1−z[∑k=0m−1(−1)k(mk)Li−k(z)]\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 - 5 years, 7 months ago

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Nice!

Jake Lai - 5 years, 7 months ago

A point to note is that Li⁡−m(z)\operatorname{Li}_{-m} (z) , when mm is a positive integer, already has a closed form in terms of rational polynomial functions.

Ishan Singh - 5 years, 1 month ago
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