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.

  • $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.