I am having doubts because it seemed too easy, and I am usually not quite sharp. I would like to know if I am wrong somewhere.

$n^n\ge n!$ and therefore ${1\over \log n^n}\le {1\over \log n!}$. Let us look at $\sum{1\over \log n^n}={\sum {1\over n\log n}}$. Let us use Cauchy condensation test: ${\sum 2^n{1\over 2^n\log 2^n}}={\sum {1\over n\log 2}}={1\over \log 2} {\sum {1\over n}}$ which diverges. Therefore $\sum{1\over \log n^n}$ diverges, and since it is smaller than the original series, so does the original series. I would like any ideas as for where I have been wrong. This exam got me really nervous and I am having troubles thinking rationally.

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    $\begingroup$ You shouldn't have any doubts at all because your proof is totally fine! Well done! $\endgroup$ – sranthrop Mar 7 '15 at 17:15
  • $\begingroup$ Thank you for your confirmation, that is really calming! :) $\endgroup$ – Meitar Abarbanel Mar 7 '15 at 17:18
  • $\begingroup$ You are very welcome :) $\endgroup$ – sranthrop Mar 7 '15 at 18:09
  • $\begingroup$ Integral test on $1/(n\log n)$ works too. $\endgroup$ – jdods Mar 7 '15 at 23:57

You might be interested to know that wolframalpha does convergence tests. It doesn't provide the proof, but it does give the solution. In this case, you are correct, the series diverges.


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