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Let G be the gamma function, and b be a constant in (-2,inf). Let

H(n, i) = G(i+1+b) * G(n-i+1+b) / [G(i+1) * G(n-i+1)]

for integers n > i > 0. Let

S(n) = \sum_{i=1}^{i=n-1} H(n, i).

Let x_ n = H(n,1) / S(n). Note x_ 2 = 1, x_ 3 = 1/2 for all b.

I am convinced that as n -> inf, x_ n -> 0 for b >= -1, and x_ n -> (-b-1)/2 for -2 < b < -1. I can prove the b >= -1 case, but not the other, except when b=-3/2. Can anyone help with a proof?

I have found recursive relationships between x_ n and S(n):

S(n+1) = (1/(n+1)) [n + 2 + 2b + 2(n+b)x_ n/n] S(n)

x_ {n+1} = (n+b)/n x_ n S(n) / S(n+1) = (n+b)(n+1)x_ n / [n(n+2+2b) + 2(n+b)x_ n]

which may be of use. One way to deal with the b >= -1 case is use the latter to relate 1/x_ {n+1} and 1/x_ n and show this tend to infinity.

For background, see section 4 of Probability Distributions on Cladograms (1996) by David Aldous In Random Discrete Structures (its available free online)


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up vote 3 down vote accepted

Using Mathematica and using reflection formulae for Gamma one finds:

x[n,b] = (b+1) n/(n+b) G[n+b+1]/G[n+2b+2] / ( G[b+1]/G[2b+2] - 2 G[n+b+1]/G[n+2b+2] )

Now, observe that for b<-1 the quotients G[n+b+1]/G[n+2b+2] tend to infinity as n->oo (this follows from Stirling's approximation). Accordingly, for such b,

x[n,b] -> (b+1) / (-2)

which is what you predicted. I don't think that b>-2 is needed.

To prove the above formula by hand (or to see why a computer can do this), you may want to have a look at the WZ method (the book A=B by Petkovsek, Wilf and Zeilberger is a wonderful and freely available introduction).

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That's great. It is straightforward to verify your answer from the recursive relation for x_n. Graham – Graham Jones Oct 18 '09 at 19:02

This looks like a problem involving hypergeometric functions. If you state your problem in standard hypergeometric notation, you may be able to look up your answer. See Abramowitz and Stegun's Handbook of Mathematical Functions or the hypergeometric function section on Wolfram's site.

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