$\displaystyle a_n = \frac{5n!}{2^n} $

I don't understand why this sequence is divergent. Any help is much appreciated.

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- Apr 21st 2008, 04:33 PMminiwheats[SOLVED] Don't understand why this sequence is divergent.
$\displaystyle a_n = \frac{5n!}{2^n} $

I don't understand why this sequence is divergent. Any help is much appreciated. - Apr 21st 2008, 04:36 PMMathstud28
- Apr 21st 2008, 04:58 PMminiwheats
Sorry, I should've rephrased my question. I don't understand how to prove that the sequence is divergent. My solutions manual shows this:

$\displaystyle a_n = \frac{5n!}{2^n} = \frac{5}{2} \times \frac{2}{2} \times \frac{3}{2} \times \frac{4}{2} ... \frac{n-1}{2} \times \frac{n}{2} \geq \frac{5}{2} \times \frac{n}{2} \ \ for\ n > 2 = \frac{5n}{4} \rightarrow \infty, so\ {a_n}\ diverges. $

But I don't understand what they did. - Apr 21st 2008, 05:01 PMMathstud28
- Apr 21st 2008, 06:09 PMminiwheats
That works too. Thanks!