Prove that (5^n)/n! approaches 0 as n approaches infinity.

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- Mar 4th 2010, 08:05 PMSlazenger3Proving limits
Prove that (5^n)/n! approaches 0 as n approaches infinity.

- Mar 4th 2010, 08:40 PMProve It
- Mar 4th 2010, 08:42 PMDrexel28
- Mar 4th 2010, 08:43 PMProve It
- Mar 4th 2010, 08:44 PMDrexel28
- Mar 4th 2010, 08:45 PMProve It
In this case it would be.

I think it is bleedingly obvious that

$\displaystyle \frac{5 \cdot 5 \cdot 5 \dots 5}{n(n - 1)(n - 2)\dots 3 \cdot 2}$

must go to $\displaystyle 0$ since the denominator is so much bigger than the numerator for large values of $\displaystyle n$. - Mar 4th 2010, 08:48 PMDrexel28
- Mar 4th 2010, 10:05 PMdavismj
A simple application of the ratio test shows that $\displaystyle \displaystyle\lim_{n\to\infty}\frac{5^n}{n!} = \displaystyle\lim_{n\to\infty}\frac{5^{n+1}n!}{5^n (n+1)!} = \displaystyle\lim_{n\to\infty}\frac{5}{n} = 0$

For reference, the ratio test says that $\displaystyle \displaystyle\lim_{n\to\infty} (a_n) = \displaystyle\lim_{n\to\infty}\bigg(\frac{a_{n+1}} {a_n}\bigg)$ - Mar 4th 2010, 11:07 PMProve It
- Mar 5th 2010, 04:00 AMDrexel28
- Mar 5th 2010, 04:07 AMProve It