Finding convergence of series using Ratio Test

Question

I have this series but Im not too sure whether my calculations are correct $$ \sum_{n=1}^\infty \left( \frac{(3n)!}{(n!)^3}\right) $$ $$ \left( \frac{(3(n+1)!}{((n+1)!)^3}\right)\cdot\left( \frac{(n!)^3}{(3n)!}\right) $$ $$ \left( \frac{(3(n+1)(3n)!}{((n+1)^3(n!)^3}\right)\cdot\left( \frac{(n!)^3}{(3n)!}\right) $$ and by simplifying I got to $$ \left( \frac{(3(n+1)}{(n+1)^3}\right) $$ $$ \left( \frac{(3n+3)}{(n^3+3n^2+3n+1)}\right) = 0 $$ so the series is convergent by Ratio test.

Answer 1

There is a mistake in your first step, indeed by ratio test we obtain

$$ \frac{(3n+3)!}{((n+1)!)^3} \frac{(n!)^3}{(3n)!}=\frac{(3n+3)(3n+2)(3n+1)}{(n+1)^3} $$

since $(3(n+1))!=(3n+3)!$.

---

To avoid ratio test we can use that

• $(3n)! =3n\cdot(3n-1)\cdot \ldots \cdot (n+1) \cdot n!\ge (n+1)^{2n} \cdot n! \ge n^{2n} \cdot n!$

and then

$$ \frac{(3n)!}{(n!)^3} \ge \left(\frac{n^n}{n!}\right)^2 \to \infty$$