Convergence of a sequence

**akolman** · August 29th 2010, 07:08 AM

Does the following converge?

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

Thanks in advance,

**Defunkt** · August 29th 2010, 07:16 AM

Recall that:

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

And use the comparison test.

**chisigma** · August 30th 2010, 11:51 AM

The so called 'Riemann zeta function' is defined as...

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

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

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

... so that is...

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

Kind regards

$\chi$ $\sigma$

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