# An identity on Polygamma

I would like to know how to prove that:

$$\psi^{(n)}(z)=(-1)^{n+1}n! \sum_{k=0}^{\infty}\frac{1}{\left ( k+z \right )^{n+1}}$$

I know that $\displaystyle \sum_{n=0}^{\infty}\frac{1}{n+z}=-\psi (z)$ but despite the differentiation I apply I cannot get the result . I get the alternating term but not the factorial.

• Try differentiating it a few times, try to see the pattern. – Akiva Weinberger Feb 18 '15 at 11:21
• That should read $\displaystyle \sum_{n={\color{red}{-\infty}}}^{\infty}\frac{1}{n+z}=-\psi (z)$, since $\displaystyle \sum_{n=0}^{\infty}\frac{1}{n+z}$ diverges $($obviously$)$. – Lucian Feb 18 '15 at 12:31

## 1 Answer

For $f(z)=\frac{1}{k+z}$ you have $f'(z) = - \frac{1}{(k+z)^2}, f''(z) = - (-2) \frac{1}{(k+z)^3}, - (-2) (-3) \frac{1}{(k+z)^4}, ...$.

From $\frac{1}{k+z} = \frac{1}{k}(1-\frac{z}{k}+(\frac{z}{k})^2+(\frac{z}{k})^3-...)= \sum_{i=0}^\infty \frac{1}{i!}(\frac{1}{k+z})^{(i)}z^i$ you obtain the result ($(i)$ means i-th derivative by $z$).