Proof verification: The convergence of $\sum_{n=2}^{\infty}{1\over log(n!)}$.

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.

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