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. $$\int_2^\infty\frac{\mathrm dn}{n\log(n)-n}=\log(\log(n)-1)\Big|_2^\infty$$
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No, any series can be bounded above by a divergent series. For example, $\sum \frac 1/n^2 < \sum 1$. – Quang Hoang Nov 04 '18 at 08:23
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What are you indicating with $\frac{n(n+1)!}2$ is $\frac n 2 (n+1)!$? – user Nov 04 '18 at 08:25
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@QuangHoang how do you think should I show the convergence of the original sum? – Euleroid Nov 04 '18 at 08:25
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1@gimusi how are the two different? – Euleroid Nov 04 '18 at 08:26
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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
$$\frac{\frac1{\log\left(\frac{n(n+1)!}2\right)}}{\frac1{n\log n}}=\frac{n\log n}{\log\left(\frac{n(n+1)!}2\right)} \to L$$
user
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@Euleroid The evaluation by Stirling help to guess that $b_n=1/n\log n$. Now to forlmalize we need to take the limit $a_n/b_n$. – user Nov 04 '18 at 09:10
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@Euleroid As you can see it is a very effective test since we can at first estimate and then just compare the limit without serach for correct inequalities. – user Nov 04 '18 at 09:14
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