# How do I solve: lim n-->inf (2n+1)!/(2n+3)!?

originally, it was the sum of the series (from 0-inf) of (-1)^n/(2n+1)!

I need to find if it's converging absolutely or conditionally so I proved it converged using the alternating series test and I thought of using the ratio test to prove that it either diverges or converges and got stuck at the part above. Any advice?

### 2 Answers

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- Tex-sLv 58 years ago
not explicit but as n increases, say to 100....

you get

21! / 23! or 1 / (202 * 203) ---> 0

- Anonymous8 years ago
(2n+3)!=(2n+3)(2n+2)(2n+1)!

(2n+1)! will cancel.

1/(2n+3)(2n+2).

That goes to 0, woo. But there's no need to prove it in multiple ways.

And technically it should be negative since (-1)^(n+1)/(-1)^(n)= -1

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