What is the serie convergence of $ \sum_{n=1}^\infty \frac{(3n)!}{n^{3n}} $

Question

I am stucked with this example:

$$ \sum_{n=1}^\infty \frac{(3n)!}{n^{3n}} $$

Im proving this using d'alembert principle

So I have

$$ \lim_{x\to \infty} \frac{\frac{(3(n+1))!}{(n+1)^{3(n+1)}}}{\frac{(3n)!}{n^{3n}}} $$

So after some calculations i get

$$ \lim_{x\to \infty} \frac{(3n)!(3n+3)(3n+2)(3n+1)(n^{3n})}{(3n)!(n+1)^{3(n+1)}} $$

$$ \lim_{x\to \infty} \frac{(3n+3)(3n+2)(3n+1)(n^{3n})}{(n+1)^{3n+3}} $$

$$ \lim_{x\to \infty} \frac{(3n+3)(3n+2)(3n+1)(n^{3n})}{n^{3n+3}+1} $$

And I don't know how to cut the powers $n^{3n}$

$$ \frac {n^{3n}}{ n^{3n} * n^3 +1} $$

I will be very thankful for every help and explanation. I hope I named this problem corectly in english cause it isnt my mother tongue.

Answer 1

The ratio simplifies to $$ \frac{(3n+3)(3n+2)(3n+1)}{(n+1)^3} \cdot \left(\frac{n}{n+1} \right)^{3n} $$ Now the first factor goes to $3^3 = 27$. For the second factor use the well-known limit $\lim_{n \to \infty} \left(\frac{n+1}{n} \right)^n = e$. Therefore the entire limit is $\frac{27}{e^3} > 1$, and the series is therefore divergent.

Answer 2

Replacing $3n=k$ and using the Stirling's approximation note that $$ \frac{k!}{(\frac {k}{3})^k} \sim_{+\infty} \frac {\sqrt{2\pi k} (\frac {k}{e})^k}{(\frac {k}{3})^k}= \sqrt{2\pi k}(\frac {3}{e})^k$$ but the series $$\sum_{k\ge 1} \sqrt{2\pi k}(\frac {3}{e})^k$$ diverges