Calculus 2. Series. Alternating Series. How do I take the limit to infinity when I have n!'s in the numerator and n's in the denominator?

Question

Calculus 2.  Series.  Alternating Series.  How do I take the limit to infinity when I have n!'s in the numerator and n's in the denominator?

anonymous · Answer

Can you give an example?

anonymous · Answer

The Sum E (n!)^2 * 3^n / (2n+1)!     I hope that kind of makes sense, the E is the sum to infinity with n= 1

anonymous · Answer

Im just not sure how to plug in infinity when I take the limit and deal with the factorials without using a different series test, the thing is that it's actually (-1)^ n so im already doing an alternating series test

anonymous · Answer

$$\sum_{n=1}^{\infty} \frac{n!^{2} 3^{n}}{(2n+1)!}$$Right?

anonymous · Answer

yes nice

anonymous · Answer

but a (-1)^n out front also

anonymous · Answer

Is there a power of x, too?

anonymous · Answer

no.. I don't think so

anonymous · Answer

It's been a long time since I've had to do these.  Hopefully someone else will come along that knows more.  Sorry.

anonymous · Answer

it's fine, thanks for trying to help!

anonymous · Answer

Hopefully someone will answer so I can find out, too!