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Determine whether the series converges $\sum_{k=1}^\infty \frac{(k!)^2}{(2k)!}$

Attempt: I used ratio test, but I guess I am making a mistake in cancelling out terms.$\lim_{k\rightarrow \infty}\frac{((k+1)!)^2}{(2(k+1))!} \frac{(2k)!}{(k!)^2}=\lim_{k\rightarrow \infty } \frac{k+1}{2}$

I am not experienced with factorials. For example, I know that $(k+1)!=k!(k+1)$, but I cant figure what $(2(k+1))!$ equals to. Help appreciated.

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    Just note that $(2k+2)! = (2k+2)(2k+1)(2k)!$.2012-10-17

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Rewrite $2(k+1)=2k+2$, instead, so that $\bigl(2(k+1)\bigr)!=(2k+2)!=(2k+2)(2k+1)(2k)!.$ Observe, too that $\frac{\bigl((k+1)!\bigr)^2}{(k!)^2}=\left(\frac{(k+1)!}{k!}\right)^2=\left(\frac{k!(k+1)}{k!}\right)^2=(k+1)^2,$ (not simply $k+1$). Hopefully that gets you as far as you need.

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    It did. Thanks.2012-10-17
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You have the quotient wrong:

$\frac{((k+1)!)^2}{(2(k+1))!} \frac{(2k)!}{(k!)^2}=\frac{(k+1)^2}{(2k+1)(2k+2)}\xrightarrow [k\to\infty]{}\frac{1}{4}$

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    @Dostre: Combining this with sos440's comment gives a good picture. You can only cancel factorials when the bases (don't know if that is a good word, but looking for the thing you take factorial of) match, but if the bases are close, you can take out the factors necessary to make them match.2012-10-17