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[[User:Idryg|Idryg]] 14:51, 28 October 2008 (UTC) | [[User:Idryg|Idryg]] 14:51, 28 October 2008 (UTC) | ||
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| + | I would try writing out the numerator and denominator and see how they compare. Another thing you will need to notice is that there are n number from 1 to n in the factorial and there are n n's in the denominator for the exponential. That means there are the same number of things getting multiplied on the top as on the bottom. if you line them up you'll see how it compares to 1/n. | ||
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| + | <math>a_n=\frac{n!}{n^n}=\frac{1*2*3*4*...*n}{n*n*n*n*...*n}</math> | ||
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| + | [[User:Jhunsber|His Awesomeness, Josh Hunsberger]] | ||
Revision as of 09:55, 28 October 2008
Okay, so the problem is:
$ a_n = \frac{n!}{n^n} $
And, it hints to compare this with $ \frac{1}{n} $.
I don't get really how to compare these two.. I mean, I can see that the denominator of $ a_n $ will grow much faster than the numerator.
I'm guessing the limit approaches 0 as n approaches infinity, but I'm not sure how to mathematically show that
$ a_n < \frac{1}{n} $
Idryg 14:51, 28 October 2008 (UTC)
I would try writing out the numerator and denominator and see how they compare. Another thing you will need to notice is that there are n number from 1 to n in the factorial and there are n n's in the denominator for the exponential. That means there are the same number of things getting multiplied on the top as on the bottom. if you line them up you'll see how it compares to 1/n.
$ a_n=\frac{n!}{n^n}=\frac{1*2*3*4*...*n}{n*n*n*n*...*n} $
