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**DBA** I need to find out if the Alternating Series is convergent or divergent by the Alternating Series Test.

One condition is to check if the limit is zero.

I have problems finding the limit, since I have a term (2k-1)! in the denominator and I do not know how to handle this...

$\displaystyle

\Sigma^{\infty}_{k = 1} (-1)^{k-1}\frac{k!}{(2k-1)!}

$

$\displaystyle

\lim_{k \to \infty}\frac{k!}{(2k-1)!}

$

How can I write the fraction, so that things cancel out?

$\displaystyle

\frac{k!}{(2k-1)!} = ?

$

For the following example I was able to write one term:

$\displaystyle

\frac{k!}{(k+1)!} = \frac{k (k-1)(k-2)(k-3)(k-4) ... (2)(1)}{(k+1)(k)(k-1)(k-2)(k-3)(k-4)....(2)(1)} = \frac{1}{k+1}

$

Can someone please let me know how to do this with

$\displaystyle

\frac{k!}{(2k-1)!}

$

Thanks!