m
m
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Writing it as: <br>
 
Writing it as: <br>
  <math> n \cdot \sum_{i=n}^1\frac{1}{i}\!<\math>  
+
  <math> n \cdot \sum_{i=n}^1\frac{1}{i}\!</math>  
<br> can help you see that the summation is the Harmonic number. I do not believe there is a closed form to this number.  
+
<br> can help you see that the summation portion is the Harmonic number. I do not believe there is a closed form to this number.  
  
 
AJ
 
AJ
 
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Revision as of 16:36, 6 October 2008

E as opposed to P


I am not entirely certain, but since $ \frac{n - i + 1}{n}\! $ is the probability of getting a different coupon for each one, souldn't the expected value be:


$ \sum_{i=1}^n\frac{n}{n - i + 1}\! $


The sum of the individual expected values should be the expected value of the sum, right? Just a thought. I am not sure, if that is right.


Perhaps, someone else could expand on this idea with more word-for-word, verbatim note-copying.


Virgil, you are right. The question is, can this be simplified to a closed form (non-summation) equation? Still trying to determine this.

Ken


Writing it as:

$  n \cdot \sum_{i=n}^1\frac{1}{i}\! $ 


can help you see that the summation portion is the Harmonic number. I do not believe there is a closed form to this number.

AJ


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