# The series $\sum_{n=2}^{\infty }\frac{1}{\log (n!)}$

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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The first two terms are undefined. – joriki Apr 12 '11 at 10:33
I guess the summation should start at $n=2$. Otherwise the series is trivially divergent... – Fabian Apr 12 '11 at 10:33
@joriki,@Fabian: Thank you, I'll fix it. – user6163 Apr 12 '11 at 10:40
@Nir: please edit your post to improve the spelling, punctuation and grammar. (Yes, people care about such things here.) Especially, please change "Convergents" to "converges" and "Delamber" to "d'Alembert" (the latter really puzzled me when I first read your message: in fact, many more people will know what you mean if you just say ratio test). – Pete L. Clark Apr 12 '11 at 15:29
@Nir: thanks! +1 for your edit (and also for posting your own solution to the problem). – Pete L. Clark Apr 12 '11 at 15:39

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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sorry for not mention that, I can't use the intergal test neither here. – user6163 Apr 12 '11 at 10:41
@Nir: so what you can use? – Fabian Apr 12 '11 at 10:42
all the other tests. – user6163 Apr 12 '11 at 10:43
Rudin gives a proof based on condensation. – lhf Apr 12 '11 at 10:45
Cauchy Condensation will now work fine instead of integral test here. – Simon Morris Apr 12 '11 at 10:46

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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That's what Fabian suggested. – lhf Apr 12 '11 at 13:45
He did not use this inequality. – user6163 Apr 12 '11 at 14:53
@lhf: Nir said "You both gave me an Idea". His solution is indeed not exactly the same as Fabian's, but he says straight out that it is inspired by it. I find it praiseworthy when OP's reason through answers given here and post their answers in their own words. – Pete L. Clark Apr 12 '11 at 15:42
@Pete, sure. Sorry for the noise. – lhf Apr 12 '11 at 18:20