I'm having some trouble proving that this sequence diverges. Can anyone suggest how to approach & proceed?
$a_n = \frac{n!}{2^n}$
I'm having some trouble proving that this sequence diverges. Can anyone suggest how to approach & proceed?
$a_n = \frac{n!}{2^n}$
Note that $n! \geq n\cdot 2^{n-2}$ (at least for large $n$)
Thus $$ a_n = \frac{n!}{2^n} \geq \frac{n2^{n-2}}{2^n} = n\cdot\frac{1}{4} \stackrel{n\to\infty}{\longrightarrow}\infty$$
Which means $a_n$ tends to $\infty$
Where $n! \geq n\cdot 2^{n-2}$ holds because
$$n! = n\cdot\underbrace{\underbrace{(n-1)}_{\geq2}\cdot\underbrace{(n-2)}_{\geq2}\dots\cdot\underbrace{3}_{\geq2}\cdot\underbrace{2}_{\geq2}}_{n-2 \text{ factors }}\cdot1$$
Note that $a_n \geq 1$ for all n. Also, $a_{n+1}/a_n=(n+1)/2$. So it's clear that $a_n$ can't converge, as otherwise $a_{n+1}/a_n$ would have limit $1$ while $(n+1)/2$ diverges.
$n!/2^n=(\frac{1}{2})(\frac{2}{2})(\frac{3}{2})\cdots (\frac{n}{2})$. Except for the first two factors, these are all at least $\frac{3}{2}$, so the product is at least $(\frac{1}{2})(\frac{2}{2})\cdot (\frac{3}{2})^{n-2}$ for $n>2$.