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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 $(2k+2)! = (2k+2)(2k+1)(2k)!$. – Sangchul Lee Oct 17 '12 at 3:05
up vote 2 down vote accepted

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. – Koba Oct 17 '12 at 3:56

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. – Ross Millikan Oct 17 '12 at 3:14
Thank you. Very detailed. – Koba Oct 17 '12 at 3:57

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