$\sum\frac{(k!)^2}{2k!}$ serie converge or diverge Does the following series diverge or converge.

$\sum\frac{(k!)^2}{2k!}$

I decide to apply the powerful ratio test.

I did

$\frac{(k+1)!^2}{2(k+1)!}$ 

which then becomes

$\frac{(k+1)!(k+1)!}{(2k+2)!}$*$\frac{2k!}{(k!)(k!)}$

I get the 

$k\rightarrow\infty$

$\frac{(k+1)(k+1)}{(2k+2)(2k+2)}$

=$\frac{1}{4}$

$\frac{1}{4}$&lt;1

convergent?
Assuming you're working with the series:

$$\sum\frac{(k!)^2}{(2k)!},$$ then you should be able to reduce the limit of the ratio to $$\lim_{k \to \infty} \frac{(k+1)(k+1)}{(2k+2)(2k+1)}$$

The limit is still $\frac 14$, and so, yes, by the ratio test, we have that the series converges.
Fernando, I took a course on series last year; they taught us how to sum EVERYTHING. Even if a series is divergent, it still may be possible to sum! Sum is $\frac{1}{3} + \frac{2 \pi}{9\sqrt{3}}$ . Remember, you must sum first, converge later.