Thread: [SOLVED] Convergence....like all the others

1. [SOLVED] Convergence....like all the others

$\displaystyle \sum_{n=1}^{\infty}\frac{(n!)^2}{2n!}$

Unlike some other that can do this one in there heads, I have to workout it out visually, I used the ration test, I believe I may have made some mistakes in simplification so I am asking for a check like always.

$\displaystyle a_n=\frac{(n!)^2}{(2n!)}$

$\displaystyle a_n+1=\frac{(n+1!)^2}{(2n+2!)}$

$\displaystyle \lim_{n\to\infty}\|\frac{(n+1!)^2}{(2n+2!)}\cdot\f rac{(2n!)}{(n!)^2}$

Doing some simplification to

$\displaystyle \lim_{n\to\infty}\frac{(n+1)^2}{(2n+2)(2n+1)}$

Multiplying out the terms in the numerator and denominator gave me a fraction with the same degree in both denominator and numerator

$\displaystyle \lim_{n\to\infty}\frac{(n^2+2n+1)}{(4n^2+6n+2)}$

I can take the ratio of the degrees of the highest powers since they the same to find limit which gives $\displaystyle L = \frac{1}{4}$ and the ratio test states, if the limit is less than 1, the series converges, is this correct?

I just got confused when wolfram, simplified the series to $\displaystyle \sum_{n=1}^{\infty}\frac{n!}{2}$ and stated it diverges, I was using wolfram for a quick check, If I did things correct or not.

2. This looks right, if your series is $\displaystyle \sum_{n = 1}^{\infty} \frac{(n!)^2}{(2n)!}$. That is, 2n factorial on the bottom as opposed to 2 times n factorial. Include the parenthesis around the 2n and you should get that the series converges to about 0.7364.

Otherwise, wolfram is right and your series diverges for sure.

3. Yes, the series was with 2n! not 2 * n!, thanks for the confirmation