# Convergence of a sequence

1. Mar 10, 2010

I am having difficulty determining whether or not the following sequence can be classified as convergent or divergent:

$$^{\infty}_{k=1}{\sum}$$$$\frac{1}{k^{ln(k)}}$$

This can be simplified to:

$$^{\infty}_{k=1}{\sum}$$$$\frac{1}{e^{{ln(k)}^{2}}}$$

Both the ratio test and root test are inconclusive (giving values of 1), while attempting the integral test doesn't work as I am unable to integrate this as a function.

Any suggestions?

Last edited: Mar 10, 2010
2. Mar 10, 2010

### l'Hôpital

If by k, you mean n, then consider the following.
$$y = e^{(ln n)^2} \rightarrow y' = y \frac{2 \ln n}{n} > 0.$$

Therefore, the terms are strictly nonincreasing.

Consider the following:

http://en.wikipedia.org/wiki/Cauchy_condensation_test

I'm sure you can do the rest.

3. Mar 10, 2010

Yes, by n I meant k.

I have actually never encountered the Cauchy condensation test until now.

I was able to finish it. Thank you very much.

4. Mar 11, 2010

### Gib Z

A comparison to a p-series would have also worked.