Is this correct? $$\sum_{n=2}^\infty\frac1{\log\left(\frac{n(n+1)!}2\right)}<\sum_{n=2}^\infty\frac{1}{\log n!}\approx\sum_{n=2}^\infty\frac1{n\log (n)-n}$$ Since the integral below does not converge, then the sum does not also converge. $$\int_2^\infty\frac{\mathrm dn}{n\log(n)-n}=\log(\log(n)-1)\Big|_2^\infty$$

  • $\begingroup$ No, any series can be bounded above by a divergent series. For example, $\sum \frac 1/n^2 < \sum 1$. $\endgroup$ – Quang Hoang Nov 4 '18 at 8:23
  • $\begingroup$ What are you indicating with $\frac{n(n+1)!}2$ is $\frac n 2 (n+1)!$? $\endgroup$ – gimusi Nov 4 '18 at 8:25
  • $\begingroup$ @QuangHoang how do you think should I show the convergence of the original sum? $\endgroup$ – Euleroid Nov 4 '18 at 8:25
  • $\begingroup$ @gimusi how are the two different? $\endgroup$ – Euleroid Nov 4 '18 at 8:26

As noticed you can't conclude for divergence with that bound (we need "our series" $\ge$ "divergent series").

By Stirling's approximation we have that

$$\frac{n(n+1)!}2 \sim \frac12n(n+1)\sqrt{2 \pi n}\left(\frac{n}{e}\right)^{n}\sim \frac k {e^n}n^{n+\frac52}$$

and therefore

$$\log\left(\frac{n(n+1)!}2\right)\sim \left(n+\frac52\right)\log n+\log k-n\sim n\log n$$

then refer to limit comparison test and show that

$$\frac{\frac1{\log\left(\frac{n(n+1)!}2\right)}}{\frac1{n\log n}}=\frac{n\log n}{\log\left(\frac{n(n+1)!}2\right)} \to L$$

  • $\begingroup$ @Euleroid It is a constant. $\endgroup$ – gimusi Nov 4 '18 at 8:57
  • $\begingroup$ Forgive me again, how do I use the test? What do I compare? $\endgroup$ – Euleroid Nov 4 '18 at 9:00
  • $\begingroup$ @Euleroid Are not you aware about Limit Comparison Test? Refer to LCT $\endgroup$ – gimusi Nov 4 '18 at 9:04
  • $\begingroup$ Sorry it wasn't clear what $b_n$ i will use. $\endgroup$ – Euleroid Nov 4 '18 at 9:07
  • $\begingroup$ @Euleroid The evaluation by Stirling help to guess that $b_n=1/n\log n$. Now to forlmalize we need to take the limit $a_n/b_n$. $\endgroup$ – gimusi Nov 4 '18 at 9:10

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