1. Determine values for convergence

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.

2. 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

$\displaystyle \chi$ $\displaystyle \sigma$

3. Originally Posted by coldfire
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.
It most definitely diverges for $\displaystyle p\le1$ since then $\displaystyle \frac{\ln(x)}{x^p}\ge\frac{1}{x^p}$. For $\displaystyle p>1$ consider the integral test.