Does $\sum_{n=2}^{\infty}\frac{n^4}{\log(1)+\log(2)+\log(3)+\cdots+\log(n)}$ converge?

Question

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).

Answer 1

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 $0$.

Answer 2

We can use the comparison test to show it diverges and we will be using the series $b_n = n^2$.

First note that $\log(n!) \approx n\log(n) - n$ by the Stirling's formula.

Then $a_n = \frac{n^4}{\log(n!)} \to \frac{n^4}{n\log(n) - n} = \frac{n^3}{\log(n) - 1} > \frac{n^3}{n} = n^2$ which diverges. Then, your series must also diverge.

Answer 3

It is easy to show that $n!\le n^n$.

Therefore, $\log(n!)\le n\log(n)$ and we have

$$\sum_{n=2}^N \frac{n^4}{\log(n!)}\ge \sum_{n=2}^N \frac{n^3}{\log(n)}\to \infty \,\,\text{as}\,\,N\to \infty$$