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

Question

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

Answer 1

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

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