# Limit of factorial in Alternating Series

• Apr 20th 2010, 12:20 PM
DBA
Limit of factorial in Alternating Series
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!
• Apr 20th 2010, 12:53 PM
Laurent
Quote:

Originally Posted by 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!

You can do the same here and get $\displaystyle \frac{k!}{(2k-1)!}=\frac{1}{(k+1)(k+2)\cdots(2k-1)}$ as soon as $\displaystyle k\leq 2k-1$, i.e. $\displaystyle k\geq 1$.