Does $\sum\limits_{k=2}^\infty\frac{1}{(\ln(k!))^2}$ converges? 
Here is my solution, but I'm unsure if it is right (the book I'm using has no solutions only "yes/no" answers):



*

*Compare with series $$\sum_{k=2}^\infty\frac{1}{(\ln(k!))}$$ which we know is divergent. 

*Set $$a_k = \frac{1}{(\ln(k!))^2}$$ and $$b_k = \frac{1}{(\ln(k!))}$$ and compare.

*$$
\lim_{k \to \infty} \frac{a_k}{b_k} = \frac{\ln(k!)}{(\ln(k!))^2} = \frac{1}{\ln(k!)} = 0
$$

*The limit is 0 so the comparison test for positive series lets us conclude that the series does not diverge so it converges.
Is the above right? Are there other methods you can use to prove convergence?
Edit: As explained in the answers, the above solution is incorrect. The fixed solution is:
$$\frac{1}{(\ln k!)^2} \leq \frac{1}{(\ln 2^{k-1})^2} = \frac{1}{(k-1)^2(\ln 2)^2} = \frac{1}{(\ln 2)^2}\cdot\frac{1}{(k-1)^2}$$
The rightmost expression is of the form $c_1\frac{1}{(k-c_2)^2}$, where $c_1$ and $c_2$ are arbitrary positive constants. So the dominoes fall:
$\frac{1}{k^2}$ converges $\implies \frac{1}{(k-1)^2}$ converges $\implies \frac{1}{(\ln 2)^2(k-1)^2}$ converges $\implies \frac{1}{(\ln 2^{k-1})^2}$ converges $\implies \frac{1}{(\ln k!)^2}$ converges.
 A: Here it is a brutal shortcut: $\log(k!)$ is a convex function by the Bohr-Mollerup theorem, hence your series is surely convergent by asymptotic comparison with $\sum_{k\geq 2}\frac{1}{k^2}$.
A: Since for $x>0$   $$\frac{x}{x+1}\le \ln(x+1)\le x$$
and Given that from the Euler Mascheroni's constant, we have $$\sum_{j=1}^{k}\frac{1}{j} \sim \gamma+\ln k~~~~as~~~k\to\infty$$ Hence, 
$$ \color{red}{ \ln(k!)=\sum_{j=1}^{k}\ln j  \ge \sum_{j=1}^{k}\frac{j-1}{j} = k -\sum_{j=1}^{k}\frac{1}{j}  \sim  k-\ln k -\gamma }\sim  k-\ln k     $$

Therefore, 
$$\sum_{k=1}^\infty\frac{1}{(\ln(k!))^2} \le C \sum_{k=1}^\infty\frac{1}{(k-\ln k)^2} \le C\sum_{k=1}^\infty\frac{1}{k^2} $$
This prove the convergence. Indeed, $$\lim_{k\to \infty }\frac{k}{k-\ln k} =\lim_{k\to \infty }\frac{1}{1-\frac{\ln k} k} = 1$$ Then for $k$ large enough we have, $$\frac{1}{(k-\ln k)^2}\le \frac{C}{k^2} $$
A: You compared your series with a divergent series. But the test is inconclusive in this case, since you got the limit $0$. Besides, you haven't said why the series $\sum \frac{1}{\log (n!)}$ is divergent. In fact, 
$$\frac{1}{\log (n!)} \simeq \frac{1}{n \log n}$$ since
$$\left(\frac{n}{e}\right)^n \le n! \le n^n $$ and so 
$$n(\log n -1) \le \log (n!) \le n \log n$$
That comparison is enough to give you convergence, since $\sum \frac{1}{(n \log n)^2}$ is convergent ( in fact already $\sum \frac{1}{n^2}$ is convergent, or even $\frac{1}{n (\log n)^2}$ is also convergent - but not $\frac{1}{n \log n}$). 
In general, if you want to prove convergence ( divergence), you may compare with another convergent(divergent) series. 
A: Let $\sum\limits_{n=2}^\infty u_n$ where $u_n=\frac{1}{(\ln(n!))^2}$. 
We have $\displaystyle\lim_{n\to\infty}\frac{u_{n+1}}{u_n}= \displaystyle\lim_{n\to\infty}\frac{(\ln(n!))^2}{(\ln((n+1)!))^2}=\displaystyle\lim_{n\to\infty}\frac{1}{(\ln(n+1))^2}=0<1$.
Then, using D'Alembert criterion , the serie $\sum\limits_{n=2}^\infty \frac{1}{(\ln(n!))^2}$ converge.
