About digamma function

Question

Show that $F(z)$ has the series expansion $\displaystyle F(z)=-\gamma + \sum_{n=2}^{\infty }(-1)^n\zeta (n)z^{n-1}$ where $\displaystyle \zeta (n)$ is the Riemann zeta function.

Answer 1

We start with the identity of the digamma function

$$ \psi(z+1) = -\gamma + \sum_{n=1}^{\infty}\frac{z}{n(n+z)}\quad z\neq 0,-1,-2,-3,\dots$$

which implies that

$$ \psi(z+1) = -\gamma + \sum_{n=1}^{\infty}\frac{z}{n^2}\left(1+\frac{z}{n}\right)^{-1} $$

$$ = -\gamma + \sum_{n=1}^{\infty}\frac{z}{n^2}\sum_{m=0}^{\infty}\frac{(-z)^m}{n^m}= -\gamma + \sum_{m=0}^{\infty}(-1)^m {z^{m+1}}\sum_{n=1}^{\infty}\frac{1}{n^{m+2}}$$

$$ = -\gamma + \sum_{m=0}^{\infty}(-1)^m \zeta(m+2){z^{m+1}} $$

$$ \implies \psi(z+1) = -\gamma + \sum_{m=2}^{\infty}(-1)^m \zeta(m){z^{m-1}}. $$