Convergence of a sequence

  • Thread starter Myriadi
  • Start date
  • #1
10
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Main Question or Discussion Point

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

[tex]^{\infty}_{k=1}{\sum}[/tex][tex]\frac{1}{k^{ln(k)}}[/tex]

This can be simplified to:

[tex]^{\infty}_{k=1}{\sum}[/tex][tex]\frac{1}{e^{{ln(k)}^{2}}}[/tex]

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:

Answers and Replies

  • #2
258
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If by k, you mean n, then consider the following.
[tex]
y = e^{(ln n)^2} \rightarrow
y' = y \frac{2 \ln n}{n} > 0.
[/tex]

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
10
0
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
Gib Z
Homework Helper
3,346
4
A comparison to a p-series would have also worked.
 

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