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So I am going to check if this series converges. $$\sum_{k=1}^\infty \frac{(2k)!}{k^{k}} $$

I used the ratio test for this, and I end up with this limit: $$\lim_{k \to \infty} \frac{(2(k+1))!\cdot{k^{k}}}{(2k)!\cdot{(k+1)}^{k+1}}$$

I have no clue on how I simplify: $$\frac{(2(k+1))!}{(2k)!} $$ I know for instance: $$\frac{(k+1)!}{k!} = k+1 $$

But the constant 2 infront of the brackets and inside the !-sign makes me a bit confused. Could someone help me out?

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

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    $\begingroup$ Argh...I should have seen this! Thanks a lot. $\endgroup$ – fejz1234 Sep 3 '16 at 12:28
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$(2(k+1))! = (2k+2)! = (2k!)(2k+1)(2k+2)$, so your last fraction simplifies to $(2k+1)(2k+2)$.

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$$\frac{(2(k+1))!}{(2k)!} = \frac{(2k+2)!}{(2k)!} = \frac{(2k+2)(2k+1)\color{red}{(2k)}\color{green}{(2k-1)}\dots}{\color{red}{(2k)}\color{green}{(2k-1)}\dots} = (2k+2)(2k+1)$$

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    $\begingroup$ Argh...I should have seen this! Thanks a lot. $\endgroup$ – fejz1234 Sep 3 '16 at 12:29

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