Does the Series ∑(1/e^(ln(k)^2)) from k=1 to Infinity Converge?

[Myriadi]

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?

 

[l'Hôpital]

If by k, you mean n, then consider the following.
<br /> y = e^{(ln n)^2} \rightarrow<br /> y&#039; = y \frac{2 \ln n}{n} &gt; 0.<br />

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.

 

[Myriadi]

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.

 

[Gib Z]

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