# Thread: Determine values for convergence

1. ## Determine values for convergence

Hi, how do I determine the values of p for which this series converges?:

$\sum_{n=1}^{\infty} \frac{ln n}{n^p}$

Thank you.

2. Let's define $\zeta(p)$ as...

$\zeta (p) = \sum_{n=1}^{\infty} \frac{1}{n^{p}}= \sum_{n=1}^{\infty} e^{-p\cdot \ln n}$ (1)

... being $p$ a complex variable. It is well known that the series (1) converges for $Re(p) >1$ so that $\zeta (p)$ is analytic in that region of complex plain. If $\zeta(p)$ is analytic, then its derivative...

$\zeta^{'} (p) = - \sum_{n=1}^{\infty} \frac{\ln n}{n^{p}}$ (2)

... exists and the series in (2) also converges for $Re(p)>1$...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by coldfire
Hi, how do I determine the values of p for which this series converges?:

$\sum_{n=1}^{\infty} \frac{ln n}{n^p}$

Thank you.
It most definitely diverges for $p\le1$ since then $\frac{\ln(x)}{x^p}\ge\frac{1}{x^p}$. For $p>1$ consider the integral test.