Prove that the sequence is convergent: $(3n)!/(n!)^3$

Question

Prove that the sequence is convergent:

$$ \frac{(3n)!}{(n!)^3} $$

I know that it will converge at $27$, but how will you prove it is a bounded sequence?

Answer 1

The result is erroneous.

Using Stirling's formula, the general term is equivalent to

$$\dfrac{(3n/e)^{3n}\sqrt{6 \pi n}}{((n/e)^{n}\sqrt{2 \pi n})^3}=k \dfrac{3^{3n}}{n}$$

with constant $k=\dfrac{\sqrt{3}}{2 \pi}$, which is divergent.

Connected : Analysis, limit of $(3n)!/(n!)^3$.

Maybe your error comes from confusion with $\dfrac{3^{3}n}{n}=27$.

Answer 2

I think what you mean is that the ratio of consecutive terms converges to $27$. Indeed, as $n\to\infty$,$$\frac{(3n+1)(3n+2)(3n+3)}{(n+1)(n+2)(n+3)}=27\underbrace{\frac{1+\frac{1}{3n}}{1+\frac2n}}_{\to1}\underbrace{\frac{1+\frac{2}{3n}}{1+\frac3n}}_{\to1}\to27.$$