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.

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- Nov 17th 2010, 12:55 PMbilalsaeedkhanConverging/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 PMPlato
If $\displaystyle n\ge 4$ then $\displaystyle n!>2^n$.

What should that tell you? - Nov 17th 2010, 01:17 PMbilalsaeedkhan
- Nov 17th 2010, 02:29 PMemakarov
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.

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I know the sequence goes to infinity