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How do I start with this series? $$\sum_{n=2}^\infty \dfrac{1}{\ln(n!)}$$ I can use any method to solve this problem. When I try using Ratio Test I get stuck with:

$$\lim\limits_{n \to \infty} \dfrac{\ln(n!)}{\ln((n+1)!)}= \infty$$

I also tried using Comparison Test where $b_n=\frac{1}{\ln(n)}$ but $b_n$ diverges which doesn't fulfill the condition. Then I tried going with Limit Comparison test but it ended up equaling to $0$.

Help is really appreciated, thank you.

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  • $\begingroup$ By the way, the limit for the ratio test equals $1$, and so the test is inconclusive. Had it been as you suggest, you could have concluded divergence. $\endgroup$ – Vincenzo Oliva Dec 6 '15 at 18:43
  • $\begingroup$ Wait, how will it equal 1 for the limit for ratio test? And which suggestion, the ratio or the comparison? $\endgroup$ – slydez Dec 6 '15 at 18:47
  • $\begingroup$ $$\lim_{n\to\infty} \frac{\log n! }{\log (n+1)! } = \lim_{n\to\infty} 1 - \frac{\log (n+1) } {\log (n+1)! } = 1- 0=1.$$ $\endgroup$ – Vincenzo Oliva Dec 6 '15 at 18:52
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This series diverges. One way to do it is to note that $n!\le n^n$ and therefore $\ln(n!)\le n\ln n$. It follows that for $n\ge 2$ we have $$\frac{1}{\ln(n!)}\ge \frac{1}{n\ln n}.$$

But the series $\sum_2^\infty \frac{1}{n\ln n}$ diverges, say by the Integral Test. So by Comparison, our series diverges.

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  • $\begingroup$ Oh! Didn't think that n!≤n^n. Thanks so much. $\endgroup$ – slydez Dec 6 '15 at 18:48
  • $\begingroup$ You are welcome. Actually, we didn't "give away" very much, since, for large $n$. $\ln(n!}$ behaves like $n\ln n-n$. $\endgroup$ – André Nicolas Dec 6 '15 at 18:58
  • $\begingroup$ See "Stirling's series" and "Stirling's formula for n!". And for large $n$, observe that $\ln (n!)$ acts like $\int_1^n \ln (x)dx=(n\ln n)-n.$ $\endgroup$ – DanielWainfleet Dec 6 '15 at 20:08

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