Taylor series of digamma

The Taylor series and integral forms of digamma function $\psi (z)$ , are given as follows:

$\begin{cases} \displaystyle \psi (1-z) = - \gamma - \sum_{k=1}^\infty \zeta (k+1) z^k & ...(1) \\ \displaystyle \psi (1-z) = - \gamma + \int_0^1 \frac{1-x^{-z}}{1-x} dx & ...(2) \end{cases}$

Where $\gamma$ is Euler-Mascheroni constant.

$\begin{aligned} \Rightarrow \sum_{k=1}^\infty \zeta (k+1) z^k & = - \int_0^1 \frac{1-x^{-z}}{1-x} dx \\ \sum_{k=1}^\infty \zeta (k+1) z^{k+1} & = - z \int_0^1 \frac{1-x^{-z}}{1-x} dx \\ \sum_{n=2}^\infty \zeta (n) z^{n} & = - \int_0^1 z \frac{x^{-z}-1}{x-1} dx \end{aligned}$

$\Rightarrow a + b + c = 1 + 0 + 1 = \boxed{2}$

$\begin{aligned} \sum_{n=2}^{\infty} \zeta(n)z^n & = \sum_{n=2}^{\infty} z^n \sum_{k=1}^{\infty} \frac{1}{k^n} \\ & = \sum_{k=1}^{\infty} \sum_{n=2}^{\infty} \left(\frac{z}{k}\right)^n \\ & = \sum_{k=1}^{\infty} \frac{\left(\frac{z}{k}\right)^2}{1 - \frac{z}{k}}\\ & = \sum_{k=1}^{\infty} \frac{z^2}{k(k-z)}\\ & = \sum_{k=1}^{\infty} \left(\frac{z}{k-z} - \frac{z}{k}\right)\\ & = \sum_{k=1}^{\infty} z \int_{0}^{1} \left(x^{k-z-1}-x^{k-1}\right)dx\\ & = \int_{0}^{1} z \sum_{k=1}^{\infty} \left(x^{k-z-1}-x^{k-1}\right) dx\\ & = -z \int_{0}^{1} \frac{x^{-z} - 1}{x - 1}dx \end{aligned}$