# Thread: Ratio Test on a Series

1. ## Ratio Test on a Series

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

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

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

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

I can't seem to simplify this any further.

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

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

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

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

I can't seem to simplify this any further.

Where did you take the squares from in your third line? Anyway, $\frac{(3n)!}{(3n+3)!}=\frac{1}{(3n+1)(3n+2)(3n+3)}$ and from here you get at once that the limit is zero and thus the series converges.

Tonio