# Finding an expression for the series

1. Dec 7, 2011

### Timebomb3750

1. The problem statement, all variables and given/known data
1+2x+(4!x^(2))/((2!)^(2))+(6!x^(3))/((3!)^(2))+(8!x^(4))/((4!)^(2))+.....

2. Relevant equations

Uhhhhh...

3. The attempt at a solution

Well, the denominator looks like it's going to be (n!)^2. And this expression appears to be using values of N≥0. But I'm not so sure as to what the numerator may look like.

2. Dec 7, 2011

### Char. Limit

Just gonna import my answer over to here...

3. Dec 7, 2011

### Timebomb3750

Yea. Sorry about the screw-up of the thread I made. Can't type today, I guess. But thanks. Makes since.

4. Dec 7, 2011

### Timebomb3750

Alright. I think I got it. The true expression should be (2n!x^(n))/((n!)^(2))

5. Dec 7, 2011

### Char. Limit

That'll do it! Now just ratio-test that and simplify, simplify, simplify. You should get your answer before too long.

6. Dec 7, 2011

### Timebomb3750

Uh, yea. Still not getting a radius of convergence of (1/4). Could be an error in the book, or I'm doing something wrong. I think I'll ask my professor about this tomorrow.

7. Dec 7, 2011

### Char. Limit

What DO you get? If you show your work, I'm sure I can help you.

8. Dec 7, 2011

### Timebomb3750

Okay. I when I do the ratio test, it comes out to be the following...

(2(n+1)!x^(n+1))/((n+1)!^(2)) * (n!^(2))/(2n!x^(n))

this appears to equal....

(xn!)/((n+1)!)

But the limit of this as n→∞ equals zero. Still don't see how I'll get an answer of (1/4).

9. Dec 7, 2011

### Char. Limit

I think the problem is that you're seeing 2n! and 2(n+1)! as 2(n!) and 2((n+1)!), when you should be using (2n)! and (2(n+1))!. At least, that's the mistake the computer program I'm using to check my work is having. Fix that and you should be fine.

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