# [SOLVED] Don't understand why this sequence is divergent.

• Apr 21st 2008, 04:33 PM
miniwheats
[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 PM
Mathstud28
Quote:

Originally Posted by miniwheats
$\displaystyle a_n = \frac{5n!}{2^n}$

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

Becasue $\displaystyle \lim_{n \to {\infty}}a_n=\lim_{n \to {\infty}}\frac{5n!}{2^n}=\infty$ this can be either shown by knowledge of rate of increase of n! vesrsus 2^n...or using a Gamma function substitution for n! and applying L'hopital's rule
• Apr 21st 2008, 04:58 PM
miniwheats
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 PM
Mathstud28
Quote:

Originally Posted by miniwheats
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.

Saying $\displaystyle \lim_{n \to {\infty}}a_n\ne{0}$ the definition of a divergent sequence isn't enough?
• Apr 21st 2008, 06:09 PM
miniwheats
That works too. Thanks!