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I'm trying to figure out whether the following series diverges or converges by using D'Alemberts (quotientcriteria), Cauchy (integral- and rootcriteria) and Leibniz convergence test for alternating series as well as direct comparisson tests.

$$ \sum_{k=2}^\infty \frac{1}{(ln(k!))^2} $$

I'm very unclear on how to parse this series. My guess is that I have to find a series which is smaller since my guess is that it does converge.

$$ ln(k!)\geq ln(k) => \frac{1}{ln(k)}\geq \frac {1}{ln(k!)}$$

But after trying the quotient criteria:

$$ \frac{(ln(k))^2}{(ln(k+1))^2} $$

I find it unclear how to continue.

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  • $\begingroup$ Use D'Alembert's ratio test $\endgroup$ – Empty Mar 12 '15 at 16:40
  • $\begingroup$ I've tried that but I don't know how to evaluate factorials within log so i get stuck at $\frac{(ln(k!))^2}{(ln(k+1)!))^2}$ $\endgroup$ – Strange Brew Mar 12 '15 at 16:43
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    $\begingroup$ First show that for $k\ge 2$ we have $k!\ge 2^{k-1}$. $\endgroup$ – André Nicolas Mar 12 '15 at 16:48
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Of course it converges. You just have to show that $\ln(k!)^2$ grows fast enough.

For example: We have $k!\ge (k/2)^{k/2}$, so for $k\ge 6$,

$$\ln(k!)^2\ge \frac{k^2}4 \ln(k/2)^2\ge \frac{k^2}{4}$$

and $\sum\limits_{k=2}^\infty \frac1{k^2}$ converges, so by comparison your series also does.

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  • $\begingroup$ Thanks, thats something I didn't know. Useful approximation! $\endgroup$ – Strange Brew Mar 12 '15 at 18:27

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