Does $\sum _{n=2}^{\infty}\frac{1}{\ln\left(n!\right)}$ converge?

Question

I spend too much time on this one,and I just can't find a way to prove anything. Please, will someone help me?

Does $$ \sum _{n=2}^{\infty}\frac{1}{\ln\left(n!\right)}$$ converge?

P.S. Does anybody knows where can I find (online) these kind of tasks (I mean, with this kind of difficulty). Thank you (in advance). :)

Answer 1

No, it does not.

It is standard to see that $\ln n! \sim_{n\to\infty} n\ln n$, so that the series with positive terms $\sum_n \frac{1}{\ln n!}$ and $\sum_n \frac{1}{n \ln n}$ have same nature by theorems of comparison.

But the latter diverges, as a Bertrand series.$^{(\dagger)}$

---

Theorem. (Bertrand series) The series $\sum_{n=2}^\infty \frac{1}{n^a(\ln n)^b}$ converges if, and only if, (i) $a>1$ or (ii) $a=1$ and $b>1$.

Answer 2

For all $n >1$, we have $n^n > n!$. Therefore, $\displaystyle \frac{1}{n^n} < \frac{1}{n!}$, and further $\displaystyle \frac{1}{\ln(n^n)} < \frac{1}{\ln(n!)}$ since $\ln(x)$ is an increasing function. By a property of logarithms, we have $\displaystyle \frac{1}{\ln(n^n)} = \frac{1}{n\ln(n)}$.

Since $\displaystyle \sum_{n=2}^\infty \frac{1}{n \ln(n)}$ can be shown to diverge with, e.g., the integral test, it follows that $\displaystyle \sum_{n=2}^\infty \frac{1}{\ln(n!)}$ diverges per the comparison test.