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I'm trying to find out whether this Series $\sum_{n=2}^{\infty } a_{n}$ converges or not when $a_{n}=\frac{1}{\log (n!)}$

I tried couple of methods, among them: d'Alembert $\frac{a_{n+1}}{a_{n}}$, Cauchy condensation test $\sum_{n=2}^{\infty } 2^{n}a_{2^n}$, and they both didn't work for me.

Edit: I can't use stirling, and integral.

Thank you

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    @Pete L. Clark: Oh Delamber mean d'Alembert. I thought it is a strange name english native speakers use for the ratio test...2011-04-12

2 Answers 2

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Hint: You can use $a_n = \frac{1}{\log n!} = \frac{1}{\sum_{k=1}^n \log k} \geq \frac{1}{n \log n}.$ Then use the Cauchy condensation test...

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    Cauchy Condensation will now wor$k$ fine instead of integral test here.2011-04-12
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You both gave me an Idea:

We know that from $n=2$

$n!< n^{n}$, so $\frac{1}{ n\log n}<\frac{1}{\log n!}$

and now from Cauchy Condensation $\frac{1}{ n\log 2}$ is obivously diverges and we're done.

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    @Pete, sure. Sorry for the noise.2011-04-12