Rule for Series

I apologize if this seems really stupid, but I've been stuck in finding the general pattern for the following series:

$$\sum_{n=1}^{\infty}\frac{2\cdot4\cdot6\cdots(2n)}{1\cdot3\cdot 5 \cdots(2n-1)}$$

The numerator is simple enough, it's just $2^nn!$. But what I'm really having trouble is finding the rule for the denominator. I've been racking my brain, but I can't come up with anything. Maybe I'm not thinking correctly? But anyways, if someone could point me in the right direction that'd be great!

• @BrianM.Scott Aak. My mistake. – Gyu Eun Lee Oct 23 '13 at 4:11

3 Answers

If you know what the numerator is, and you know what the numerator times the denominator is, then you know what the denominator is.

• Could you elaborate as to what "the numerator times the denominator" is? I'm not sure exactly what you mean. – user95297 Oct 23 '13 at 5:21
• By "the numerator times the denominator" I mean the product of the numerator and the denominator. The quantity $N\times D$ where $N$ is the numerator and $D$ is the denominator. – bof Oct 23 '13 at 5:28
• Maybe I'm not understanding this right? Like I know for the first term, the product of NxD = 2 and N = 2, so D = 1. For the second term, the product of NxD = 24 and N = 8, so D = 3. I think what I'm really trying to say (if I'm understand you correctly) is how this will help find the rule for the denom. – user95297 Oct 23 '13 at 5:32
• The numerator is $N=2\cdot4\cdot6\cdots(2n)$. The denominator is $D=1\cdot3\cdot5\cdots(2n-1)$. If you know what $N$ is and you know what $ND$ is, then you know what $D$ is. You stated in your question that $N=2^nn!$. The formula for $ND$ is even more obvious. – bof Oct 23 '13 at 6:15
• For example, if $N=1\cdot3\cdot5\cdot7$ and $D=2\cdot4\cdot6\cdot8$, then $ND=1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8=8!$. – bof Oct 23 '13 at 6:26

Hint: each factor in the numerator is greater than the corresponding factor in the denominator. What does that suggest to you about the terms of this infinite series?

• That each subsequent term will always be greater than the previous. But I don't see how that helps, or am I still missing something? – user95297 Oct 23 '13 at 5:28
• So does the series converge? – Robert Israel Oct 23 '13 at 6:42

$\displaystyle{\large{2^{2n}\left(n!\right)^{2} \over \left(2n\right)!}}$