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.
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.