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FLMPotD - Ultrafilter! Waffle!
March 31st, 2011
01:10 pm


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Compute \sum_{k=0}^n (-1)^k \binom{n}{k}/(k+1).

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(6 comments | Leave a comment)

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From: darkerline
Date:March 31st, 2011 11:46 pm (UTC)
It turns out Wolfram|Alpha can do this one.
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From: cesium12
Date:April 2nd, 2011 04:10 pm (UTC)
You mean Bieber|Beta?
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From: sergeibernstein
Date:April 1st, 2011 12:14 am (UTC)
Ooh. An FLMPotD that I can solve. :-D
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From: limitedcake
Date:April 1st, 2011 01:26 pm (UTC)
That was fun!
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From: oxeador
Date:April 3rd, 2011 05:38 am (UTC)
For once I am allowing you to use the letter L.

I solved it by taking the binomial formula and integrating it. Is there a simple, combinatorial argument?
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From: ultrawaffle
Date:April 3rd, 2011 09:31 pm (UTC)
I was hoping someone would come up with a nice combinatorial argument! I know of two ways of doing it. The simplest way is to rewrite \binom{n}{k}/(k+1) as \binom{n+1}{k+1}/(n+1) and using the binomial theorem (since the denominators no longer depend on k). You can turn this into a combinatorial argument, but it seems kind of clunky to me and I don't see how to make it elegant. The other way, which sounds like basically the way you did it, is how I originally came across this. I was idly thinking about things like "if you take n random numbers (with respect to some distribution), what is the expected value of the largest one?" This sum came out of doing an integral to compute the following: if you pick n numbers in [1,\infty) with respect to the probability density function 1/x^2, what is the probability that the last one is the largest of them? In this formulation, of course, the answer is obvious.
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