# Prove that $\sum_{n=1}^{\infty}x^n\frac{(n!)^3}{(3n)!}$ converges when $|x|$ < 27 and diverges when $|x| > 27$

This is a homework question that I am stuck on... I am not sure which test to use to prove this statement. If someone could let me know at least which test to use to push me in the right direction that would be great.

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Hint: Apply the ratio test. Why the ratio test? Any time there are factorial terms, and powers, it works very well. Recall that if $$r=\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}$$ then $\sum_{n=1}^\infty a_n$ converges absolutely if $r<1$ and diverges if $r>1$.
Hint 2: To get you started on the limit, notice that $$r=\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=\lim_{n\rightarrow \infty} \frac{x^{n+1}(n+1)!^3}{(3n+3)!}\biggr/ \frac{x^{n}(n!)^3}{(3n)!}$$ $$= \lim_{n\rightarrow \infty}\frac{x^{n+1}(n+1)!^3(3n)!}{x^n(n!)^3(3n+3)!}=x\lim_{n\rightarrow \infty}\frac{(n+1)^3}{(3n+3)(3n+2)(3n+1)}.$$
Last step: This limit is $\frac{1}{27}$ so we see that $r=\frac{x}{27}$. From this when is $r>1$ and when is $r<1$?
@LoganSerman: No problem. As a wide rule the ratio test works well for things like $n!$ and $x^n$, whereas the root test works well for things like $x^n$ and $n^n$. – Eric Naslund Nov 28 '11 at 18:17