Polylogarithm recursion identity

I recently discovered a recursive identity for Lim(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.

Lim(z)=z1z[1+k=0m1(mk)Lik(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=1zn(n+1)m\displaystyle \sum_{n=1}^{\infty} z^n(n+1)^m. We have

n=1zn(n+1)m=n=1znk=0m(mk)nk=n=1znnm+k=0m1(mk)n=1znnk=Lim(z)+k=0m1(mk)Lik(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=1zn(n+1)m=1zn=2znnm=1zLim(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

Lim(z)+k=0m1(mk)Lik(z)=1zLim(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

Lim(z)=z1z[1+k=0m1(mk)Lik(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.

Note by Jake Lai
3 years, 11 months ago

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Similarly Lim(z)=(1)m1z[k=0m1(1)k(mk)Lik(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 - 3 years, 11 months ago

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Nice!

Jake Lai - 3 years, 11 months ago

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

Ishan Singh - 3 years, 5 months ago

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