Square difference formula for polylgoarithm proof

$$\mathrm{Li}_{\,n}(-z) + \mathrm{Li}_{\,n}(z) = 2^{1-n} \,\mathrm{Li}_{\,n}(z^2) $$

$$\textit{proof}$$

As usual we write the series representation of the LHS

$$\sum_{k=1}^\infty \frac{z^k}{k^n}+\sum_{k=1}^\infty \frac{(-z)^k}{k^n}$$

Listing the first few terms

$$z+\frac{z^2}{2^n}+\frac{z^3}{3^n}+\cdots +\left(-z+\frac{z^2}{2^n}-\frac{z^3}{3^n}+\cdots \right) $$

The odd terms will cancel

$$2\frac{z^2}{2^n}+2\frac{z^4}{4^n}+2\frac{z^6}{6^n}+\cdots $$

Take \( 2^{1-n} \) as a common factor

$$2^{1-n} \left( z^{2}+\frac{(z^2)^2}{2^n}+\frac{(z^2)^3}{3^n}+ \cdots \right)=2^{1-n} \sum_{k = 1}^\infty \frac{(z^2)^{k}}{k^n} = 2^{1-n} \, \mathrm{Li}_{\, n}(z^2) \, $$

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