Ratio Test

[Blitze105]

Hello,
at the moment I have a series... The constant in the series is:

C = (2n)! / (n!)^2

With that information the ratio test looks like this:

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

From this point, i move on to simplify...

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

I'm just not able to simplify correctly.
Any help at this point would be appreciated

 

[BigGlenntheHeavy]

$\displaystyle \lim_{n\to \infty}\bigg|\frac{[2n+2]!x^{n+1}}{[(n+1)!]^{2}} \ * \ \frac{(n!)^{2}}{x^{n}(2n)!}\bigg| \ < \ 1$

$\displaystyle = \ |x|\lim_{n\to\infty}\bigg|\frac{(2n+2)(2n+1)(2n)!}{[(n+1)(n!)]^{2}} \ * \ \frac{(n!)^{2}}{(2n)!}\bigg| \ < \ 1$

$\displaystyle = \ 2|x|\lim_{n\to\infty}\bigg|\frac{(n+1)(2n+1)}{(n+1)^{2}(n!)^{2}} \ * \ \frac{(n!)^{2}}{1}\bigg| \ < \ 1$

$\displaystyle = \ 2|x|\lim_{n\to\infty}\bigg|\frac{(n+1)(2n+1)}{(n+1)^{2}}\bigg| \ < \ 1$

$\displaystyle = \ 2|x|\lim_{n\to\infty}\bigg|\frac{(2n+1)}{(n+1)}\bigg| \ < \ 1$

$\displaystyle = \ 2|x|(2) \ < \ 1, \ 4|x| \ < \ 1, \ ergo, \ |x| \ < \ \frac{1}{4}, \ converges \ on \ -1/4 \ < \ x \ < \ 1/4$

$\displaystyle I'll \ leave \ it \ to \ you \ to \ find \ if \ the \ endpoints \ converge \ or \ diverge.$