## Convergence of a sequence

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?
 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.
 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.

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## Convergence of a sequence

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

 Tags converge, diverge, e^-ln(k)^2, k^-ln(k)