The convergence of $\sum_{n=2}^{\infty} {1\over \ln(n!)}$ Check the convergence of $\sum_{n=2}^{\infty} {1\over \ln(n!)}$.
I tried D'Alembert's test... Cauchy's test seems too intricate... I can't seem to understand what I should do here...
 A: Hint:
$$\ln n! = \ln n + \ln (n-1) + \cdots + \ln 2 + \ln 1 \le n\ln n$$
so
$$\frac{1}{\ln n!} \ge \frac{1}{n\ln n}$$
Does
$$\sum_{n=2}^{\infty} \frac{1}{n\ln n}$$
converge?
Hint 2:
If you can't use the integral test, determining the convergence/divergence of $\sum\frac{1}{n\ln n}$ is somewhat difficult.  Here's a hint on how you can do it:
$$\sum_{n=2}^{2^{k+1}-1}a_n = \sum_{n=1}^{k}\sum_{m=2^n}^{2^{n+1}-1}a_m$$
$$\sum_{m=2^n}^{2^{n+1}-1} \frac{1}{m\ln m} > \left(2^{n+1}-1 - 2^n + 1\right) \left(\frac{1}{(2^{n+1})\ln2^{n+1}}\right) = \frac{1}{2\ln2}\frac{1}{n+1}$$
Notice that this is similar to how we usually prove that the harmonic series diverges.
A: Notice that for all $N > 2$,
$$\sum_{n=2}^N \frac{1}{\ln n!} > \sum_{n=2}^N \frac{1}{n \cdot \ln n}$$
Now what do you know about the convergence of that second series?
A: Hint: $$\ln(n^n) \geq \ln(n!) \\ \implies \frac{1}{\ln(n^n)} \leq \frac{1}{\ln(n!)}$$ Now consider $$\int_{2}^\infty \frac{1}{\ln(x^x)}dx = \int_{2}^\infty \frac{1}{x\ln(x)}dx$$
A: To show that $\sum \frac{1}{\log(n!)}$ diverges it suffices to show that $\sum \frac{1}{n\log(n)}$ diverges as the other answers have shown.
Here is an elementary proof that $\sum_{n=2}^N\frac{1}{n\log n}$ diverges (since OP said he did not know integrals). We start with a little lemma:

If $a_n\geq 0$ is a sequence and $s_n = \sum_{i=1}^n a_i$ then 
  $$\log\left(\frac{s_N}{s_{1}}\right) \leq \sum_{n=2}^N\frac{a_n}{s_{n-1}}$$

Proof: We have
$$\log\left(\frac{s_n}{s_{n-1}}\right) = \log\left(1 + \frac{a_n}{s_{n-1}}\right) \leq \frac{a_n}{s_{n-1}}$$
since $\log(1+x)\leq x$ for all $x\geq 0$. By summing over $n=2,3,\ldots,N$ we get
$$\log\left(\frac{s_N}{s_{1}}\right) \leq \sum_{n=2}^N\frac{a_n}{s_{n-1}}$$
We first apply the lemma above to $a_n=1$ to obtain
$$\log\left(N\right) \leq \sum_{n=1}^{N-1}\frac{1}{n}$$
We then proceed to apply the lemma to $a_n = \frac{1}{n}$ to find
$$\log\left(\sum_{i=1}^N \frac{1}{i}\right) \leq \sum_{n=2}^N\frac{1}{n\sum_{i=1}^{n-1}\frac{1}{i}} \leq \sum_{n=2}^N\frac{1}{n\log n}$$
where we have used the inequality derived above in the last step. Finally since the harmonic series diverges so does the logarithm and it follows that $\sum \frac{1}{n\log n}$ diverges.
$\mathbf{Remark:}$ By continuing to apply the lemma above to $a_n = \frac{1}{n\log n}$ (and so on), the method above can be used to show that $\sum \frac{1}{n\log(n)\log(\log(n))\ldots \log^{(k)}(n)}$ diverges for any $k$. However, this result follows much more simply from the integral test.
