Ratio test cancellation trouble $$\sum\limits_{n=8}^{\infty}\frac{6^n}{(2n)!} $$
Can someone walk me through the cancellation of numbers in this ratio test problem? I seem to be forgetting something and its leading me to the incorrect denominator. I keep getting $(2n+1)!$.

Here are the steps
$$ \lim\limits_{n\to\infty}\left|\frac{ \frac{6^{n+1}}{(2(n+1))!} }{\frac{6^n}{(2n)!}}\right|= \lim\limits_{n\to\infty}\left|\frac{ \frac{6\cdot 6^{n}}{(2n+2)!}}{\frac{6^n}{(2n)!}}\right| $$
$$= \lim\limits_{n\to\infty}\left|\frac{6\cdot 6^{n}\cdot (2n)!}{6^n\cdot (2n+2)!}\right| = \lim\limits_{n\to\infty}\left|\frac{6\cdot (2n)!}{(2n+2)!}\right| $$
$$= \lim\limits_{n\to\infty}\left|\frac{6\cdot (2n)!}{(2n+2)(2n+1)(2n)!}\right| $$
$$= \lim\limits_{n\to\infty}\left|\frac{6}{(2n+2)(2n+1)}\right|=0 $$
Therefore by the ratio test, this series is absolutely convergent.
