# Converging/Diverging Sequence

• Nov 17th 2010, 12:55 PM
bilalsaeedkhan
Converging/Diverging Sequence
Prove whether the following sequence converges or diverges.

$\displaystyle \{\frac{n!}{2^n} \}\limits^\infty_{n=0}$

Since it is a sequence I cannot use any of the tests used for series. Any help would be appreciated.
• Nov 17th 2010, 01:12 PM
Plato
If $\displaystyle n\ge 4$ then $\displaystyle n!>2^n$.
What should that tell you?
• Nov 17th 2010, 01:17 PM
bilalsaeedkhan
Quote:

Originally Posted by Plato
If $\displaystyle n\ge 4$ then $\displaystyle n!>2^n$.
What should that tell you?

Yeah I know the sequence goes to infinity and that factorials grow much faster than exponentials but is there some other way to prove it? Like taking a limit or is this the only way?
• Nov 17th 2010, 02:29 PM
emakarov
The fact that f(n) > g(n) does not determine whether $\displaystyle \lim_{n\to\infty}f(n)/g(n)$ exists. However, it is easy to show that $\displaystyle \frac{1}{2}\cdot\frac{2}{2}\cdot\ldots\cdot \frac{n}{2}$ becomes arbitrarily large.

Quote:

I know the sequence goes to infinity
This is a synonym for "diverges".