# Convergence of a sequence

• Aug 29th 2010, 07:08 AM
akolman
Convergence of a sequence
Does the following converge?

$\displaystyle \sum_{n=1}^{\infty}\frac{\log n}{n^2}$

• Aug 29th 2010, 07:16 AM
Defunkt
Recall that:

1) $\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^p}$ converges for $\displaystyle p > 1$
2) $\displaystyle \displaystyle \lim_{n \to \infty} \frac{logn}{n^{\epsilon}} = 0$ for every $\displaystyle \epsilon > 0$

And use the comparison test.
• Aug 30th 2010, 11:51 AM
chisigma
The so called 'Riemann zeta function' is defined as...

$\displaystyle \displaystyle \zeta (s) = \sum_{n=1}^{\infty} \frac{1}{n^{s}}$ , $\displaystyle \Re (s) > 1$ (1)

In You derive the series (1) 'term by term' You obtain...

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

... so that is...

$\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{\ln n}{n^{2}} = - \zeta^{'} (2)$ (3)

Kind regards

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