Determining the type of convergence of a series

The specific homework problem is $\displaystyle\sum\limits_{n=0}^\infty \frac{2\cdot 4\cdot 6\cdots2n}{n!}$. It is problem #29 from the section 11.6 exercises (pg. 737) from Single Variable Calculus: Early Transcendentals - 4th Edition by James Stewart. What we have to do is determine the type of convergence that occurs, i.e. absolute convergence, conditional convergence, or divergence. I understand that the test that I will have to do to solve this problem will probably be the ratio test. The problem that I am having in doing this is identifying the expression in the numerator of the problem. I think it is a factorial of some sorts, where it starts at $2n$ and then decreases by $2$ every single time instead of $1$. I do not know how to write this expression in the form of some sort of function. So my question is: is there any way to write a factorial that decreases by $2$ instead of $1$ or do I just have the wrong expression entirely for the numerator.

EDIT: Solution: To show that I did not just blatantly copy the answer provided by savick01, I will explain how s/he got it. What happened is that, after having pulled out a $2$ from each value in the expression, it became of the form $2(1)\cdot2(2)\cdot2(3)\cdots2(n)$. Further simplification of this, after considering the fact that the values in the parenthesis form the factorial and that there are $n$ factors in a factorial, results in $2^n\cdot n!$.

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Isn't it just $n! \cdot 2^n$ ? – savick01 Feb 25 '12 at 17:09
It is also denoted by $(2n)!!$, which is defined also for odd $m$: $m!! = m \cdot (m-2) \cdot \ldots \cdot 1$. – savick01 Feb 25 '12 at 17:11
@savick01 Thank you very much, I didn't realize that I could factor things out of a factorial. – Amndeep7 Feb 25 '12 at 17:16
The standard "sum" for the divergent series is $-1$. – robjohn Feb 25 '12 at 19:39
I find the "$!!$" notation obnoxious because it looks like an iterated factorial. – Michael Hardy Feb 25 '12 at 22:24