TITLE: Does $\sum_{n=2}^{\infty}\frac{n^4}{\log(1)+\log(2)+\log(3)+\cdots+\log(n)}$ converge?
QUESTION [8 upvotes]: I can't find a way to test the convergence/divergence of this series:
$$ \sum_{n=2}^{\infty}\frac{n^4}{\log(1)+\log(2)+\log(3)+\cdots+\log(n)} $$
I tried the Cauchy method but in order to make the logarithms more manageable I grouped them all (so $\log(1)+\log(2)+\log(3)+...+\log(n)=\log(n!)$. The problem is, I don't know how to differentiate that when I need to. So I'd be grateful for some help if someone can think of a different way or just a way to improve mine (using the Cauchy method somehow so that it works).

REPLY [11 votes]: Without Stirling.
Note that the denominator is $$
\sum_{k=1}^n \log k \leq \sum_{k=1}^n \log n = n\log n
$$
from which you can lower bound the general term of your series by $$a_n\stackrel{\rm def}{=} \frac{n^4}{\log 1+\log 2+\dots+\log n} \geq \frac{n^3}{\log n}\xrightarrow[n\to\infty]{} \infty$$
and therefore the series $\sum_n a_n$ diverges, as its general term does not even go to $0$.

REPLY [7 votes]: It is easy to show that $n!\le n^n$.
Therefore, $\log(n!)\le n\log(n)$ and we have
$$\sum_{n=2}^N \frac{n^4}{\log(n!)}\ge \sum_{n=2}^N \frac{n^3}{\log(n)}\to \infty \,\,\text{as}\,\,N\to \infty$$