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And, it hints to compare this with <math>\frac{1}{n}</math>. | And, it hints to compare this with <math>\frac{1}{n}</math>. | ||
| − | I don't get really how to compare these two.. I mean, I can see that the denominator of <math>a_n</math> 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 <math>a_n < \frac{1}{n}</math> | + | I don't get really how to compare these two.. I mean, I can see that the denominator of <math>a_n</math> will grow much faster than the numerator. |
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| + | I'm guessing the limit approaches 0 as n approaches infinity, but I'm not sure how to mathematically show that | ||
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| + | <math>a_n < \frac{1}{n}</math> | ||
Revision as of 09:50, 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} $
