examine the convergence or divergence of (summation) (log(base n) n!)/(n^3)
can we estimate n! with the help of stirlings formula and then do the question
Claim: $\displaystyle \log_{n}(n!) \le n$
You can prove this by induction if needed, but observe
$\displaystyle \log_{n}(n!)=\log_{n}(n)+\log_{n}(n-1)+\log_{n}(n-2)+...\log_{n}(2)+\log_{n}(1) \le \underbrace{1+1+1+...+1+1}_{n \text{ times}}=n$
Or if you belive that $\displaystyle n! \le n^n$ and since the log function is monotonic increaseing you get
$\displaystyle \log_{n}(n!) \le \log_{n}(n^n)=n$
But either way you end up with
$\displaystyle \sum_{n=1}^{\infty}\frac{\log_{n}(n!)}{n^3} \le \sum_{n=1}^{\infty}\frac{n}{n^3}=\sum_{n=1}^{\inft y}\frac{1}{n^2}$
And we know that this converges.