# Simplifying a series with the ratio test

1. Feb 24, 2014

### MathewsMD

1. The problem statement, all variables and given/known data
Determine if the following series is divergent or convergent:

$∑_1^∞ \frac {(2)(4)(6)...(2n)}{n!}$

2. The attempt at a solution

I understand this can be simplified to:

$∑_1^∞ \frac {(2^n)(n!)}{n!}$
This can easily be seen to be divergent.

But when I do $lim_{n→∞}\frac {a_{n+1}}{a_{n}} = lim_{n→∞}\frac {1}{n}$ after simplifying. Does anyone know what I may be doing wrong or why this is not appropriate? Is there any restriction on when the ratio test can be used? Thank you! :)

2. Feb 24, 2014

### AlephZero

How did you get that? (It is wrong).

3. Feb 24, 2014

### MathewsMD

I know $a_n = \frac {(2)(4)(6)...(2n)}{n!}$
So $\frac {a_{n+1}}{a_n} = [\frac {(2)(4)(6)...(2(n+1))}{(n+1)!}] [\frac {n!}{(2)(4)(6)...(2n)}]$
This simplifies to: $\frac {2(n+1)}{2n} \frac {1}{n+1}$

Now, when the limit is taken as n→∞, then $\frac {2(n+1)}{2n} = 1$ and you're left with $lim_{n→∞}\frac {1}{n+1}$ I just simplified this to $\frac {1}{n}$. Does anyone mind telling me what I did wrong here?

4. Feb 24, 2014

### Dick

No, the ratio is 2(n+1)/(n+1)=2. Where did the extra n come from??

5. Feb 24, 2014

### MathewsMD

ahh i forgot to cancel a 2n

6. Feb 25, 2014

### Ray Vickson

Why bother with the ratio test? The nth term of the series equals $2^n$ (as you have already noted). Since $2^n$ does not go to 0 as $n \to \infty$ the series cannot possibly converge.

7. Feb 26, 2014

### MathewsMD

Yep, I just didn't know why using a different test that I wasn't getting the same solution. Problem's solved.