# Converge or Diverge

1. Mar 29, 2012

### meson0731

1. The problem statement, all variables and given/known data

summation 1 to infinity (2*4*6*8*...*(2n)) / (n!)

2. Relevant equations

3. The attempt at a solution

I'm a little confused on the "..." part... I used the ratio test and the numbers obviously cancel out, but im not sure if the (2n+2)/(2n) part cancel out to just make (2n+2) on the top. This would make the limit equal to 2 and it would diverge. But im not entirely sure if thats right...

2. Mar 29, 2012

### jgens

It looks like you want to know if the following sum converges or diverges:
$$\sum_{n=1}^{\infty}\frac{2^n n!}{n!}=\sum_{n=1}^{\infty}2^n$$
That should be pretty easy.

3. Mar 29, 2012

### meson0731

I don't see how you rewrote it like that. Would my method of using the ratio test work? And if so how exactly would the top partion cancel out.

4. Mar 29, 2012

### jgens

Try thinking about it. The ratio test works but is overkill for this problem.

The ratio test works, but you messed up calculating it. (Hint: The ratio is 2).

5. Mar 29, 2012

### Robert1986

You said you were confused about the "..." part, so I just thought that I would mention that the $n^{th}$ term in the sum is $(2*4*\cdots*(2n))/n!$.

Perhaps you knew this, but since you said you were confused by the "..." and jgens simplification, I thought you might be overlooking this.

6. Mar 29, 2012

### meson0731

When i did the ratio test i did get 2 for the limit. Does this mean that I did it right?

7. Mar 29, 2012

### jgens

Depends. In the OP you wrote something about $\lim_{n \to \infty}\frac{2n+2}{2n} = 2$. In this case the limit is correct but the ratio is not. You should get $\lim_{n \to \infty} 2 = 2$.

8. Mar 29, 2012

### meson0731

I just realized that I wrote it wrong . I got lim as x > infinity of (2n+2)/(n+1) which is 2.

9. Mar 29, 2012

### jgens

Then you solved the problem correctly :)

10. Mar 29, 2012

### Robert1986

Wait, what? The limit doesn't look right to me...

$$\lim_{n \to \infty}\frac{2n+2}{2n} = \lim_{n \to \infty}\frac{2n}{2n} + \lim_{n \to \infty}\frac{2}{2n} = 1 + \lim_{n \to \infty}\frac{1}{n} = 1$$

Right?

11. Mar 29, 2012