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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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    The first two terms are undefined.2011-04-12
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    I guess the summation should start at $n=2$. Otherwise the series is trivially divergent...2011-04-12
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    @joriki,@Fabian: Thank you, I'll fix it.2011-04-12
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    @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**).2011-04-12
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    @Nir: thanks! +1 for your edit (and also for posting your own solution to the problem).2011-04-12
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    @Pete: sure! thank you for correcting me!2011-04-12
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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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    sorry for not mention that, I can't use the intergal test neither here.2011-04-12
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    @Nir: so what you can use?2011-04-12
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    all the other tests.2011-04-12
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    Rudin gives a proof based on condensation.2011-04-12
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    Cauchy Condensation will now work 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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    That's what Fabian suggested.2011-04-12
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    He did not use this inequality.2011-04-12
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    @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.2011-04-12
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    @Pete, sure. Sorry for the noise.2011-04-12