Let G be the [gamma function](http://en.wikipedia.org/wiki/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)


Graham



Limit of sequence involving gamma functions
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).


