Convergence/divergence of the sum $\sum_{n=2}^\infty 1/ \ln(n!) $
By Emma Terry •
Is the sum
$$ \sum_{n =2}^{\infty} \frac{1}{\ln n!} $$
convergent or divergent? I have tried different methods and it doesn't work. Perhaps comparing with a divergent series will work? I'm thinking it is divergent.
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$\begingroup$It is clear that $$n! = 1 \cdot 2 \cdot \dots \cdot n \le n \cdot n \cdot \dots \cdot n = n^n$$ and since $\log$ is increasing, $\ln(n!) \le \ln(n^n) = n \ln n$. Taking the reciprocal, this gives $$\frac{1}{\ln(n!)} \ge \frac{1}{n \ln n}.$$ But by comparing to the integral $$\int_2^\infty \frac{dx}{x \ln x} = \int_2^\infty \frac{(\ln x)'}{\ln x} dx = [ \ln(\ln x)]_2^\infty = + \infty,$$ the series $\sum \frac{1}{n \ln n}$ is divergent, thus $\sum \frac{1}{\ln(n!)}$ is divergent too.
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