$$\sum_{n=2}^\infty \frac{1}{\ln(n!)}$$
I tried by comparing it to $\sum_{n=1}^\infty \frac{1}{n}$ but i seem to fail.
I think I need to compare with series that are smaller and diverge. Help.
$$\sum_{n=2}^\infty \frac{1}{\ln(n!)}$$
I tried by comparing it to $\sum_{n=1}^\infty \frac{1}{n}$ but i seem to fail.
I think I need to compare with series that are smaller and diverge. Help.
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Hint. One may observe that $$ n! \leq n^n,\quad n\geq2, $$ giving $$ \ln(n!) \leq \ln(n^n)=n\ln n $$ and, for $N\geq2$,
then let $N \to \infty$ to conclude.