Is this correct? Convergence of $\sum_{n=2}^\infty\frac1{\log\left(\frac{n(n+1)!}2\right)}$

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. $$\int_2^\infty\frac{\mathrm dn}{n\log(n)-n}=\log(\log(n)-1)\Big|_2^\infty$$

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

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

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