Originally Posted by

**adkinsjr** The series is $\displaystyle \Sigma_{n=0}^{\infty}\frac{(n!)}{(3n)!}$

$\displaystyle \lim_{n->\infty}\mid\frac{a_{n+1}}{a_n}\mid<1 \rightarrow \Sigma a_n$ converges. If it's less than 1, then the series diverges; if it's equal to 1, then the test is inconclusive.

$\displaystyle \lim_{n->\infty}\left(\frac{[(n+1)!]^2}{[3(n+1)]!}\right)\left(\frac{(3n)!}{(n!)^2}\right)$

$\displaystyle =\lim_{n->\infty}\frac{(n+1)(3n)!}{[3(n+1)]!}$

I can't seem to simplify this any further.