Hi, how do I determine the values of p for which this series converges?:
$\displaystyle \sum_{n=1}^{\infty} \frac{ln n}{n^p}$
Thank you.
Let's define $\displaystyle \zeta(p)$ as...
$\displaystyle \zeta (p) = \sum_{n=1}^{\infty} \frac{1}{n^{p}}= \sum_{n=1}^{\infty} e^{-p\cdot \ln n}$ (1)
... being $\displaystyle p$ a complex variable. It is well known that the series (1) converges for $\displaystyle Re(p) >1$ so that $\displaystyle \zeta (p)$ is analytic in that region of complex plain. If $\displaystyle \zeta(p)$ is analytic, then its derivative...
$\displaystyle \zeta^{'} (p) = - \sum_{n=1}^{\infty} \frac{\ln n}{n^{p}}$ (2)
... exists and the series in (2) also converges for $\displaystyle Re(p)>1$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$