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<math>a_n = \frac{n!}{n^n}</math>
 
<math>a_n = \frac{n!}{n^n}</math>
  
And, it hints to compar 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>

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} $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood