I was solving practice problems for my upcoming calculus 1 final and came across this problem. I'm honestly still a little lost about the series and ratio tests.
The problem itself is $$\sum_{n=1}^∞\frac{2\cdot 4\cdot 6\cdot.....(2n)}{n!}$$
I identified $$a_n = \frac{2\cdot 4\cdot 6\cdot \ldots (2n)}{n!}$$ but I'm not sure if a $$a_{n+1} = \frac{2\cdot 4\cdot 6\cdot \ldots (2n)^{n+1}}{(n+1)!}$$ or something else? Please help evaluate the limit using, $$\left|\frac{a_{n+1}}{a_n}\right|.$$ Thank you in advance!
$$ a_n=\frac{2\cdot4\cdot6\cdots 2n}{n!}=\frac{2^nn!}{n!}=2^n $$ hence $$ \frac{a_{n+1}}{a_n}=2 $$ so the series diverges (but it was clearer just by looking at $a_n$ itself).
If you don't want to factor the 2 in the numerator, just write $$ a_{n+1}=\frac{2\cdot4\cdot6\cdots 2n\cdot (2(n+1))}{(n+1)!} =\frac{2\cdot4\cdot6\cdots 2n}{n!}\frac{2(n+1)}{n+1} =2a_n $$