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# Combinatorial identity from the Kingman coalescent

The following appears to be true. Let $n \geq 1$ be an integer. Then $$\sum_{i=1}^n \frac{n! (n-1)! }{(n-i)! (n+i+1)!}i (i+1) (2i+1) = 1.$$ This cropped up when doing some calculations in coalescent theory. I found the distribution of the time of a particular event as a mixture of exponentials. I can calculate the weights numerically and the weights appear to be the terms in the above sum.

I also noticed:

• The last term ($i=n$) is the reciprocal of a Catalan number.

• The sum of alternate terms ($i$ odd or $i$ even) appears to be always 1/2.

• $i (i+1) (2i+1) / 6$ is the sum of the first $i$ squares.

Something is going on that I don't understand! Can anyone shed any light?

• Options
1.

You could ask on MathOverflow. Or, if you'd rather not, someone could ask on your behalf. I expect that such a hypergeometric identity would succumb to Wilf-Zeilberger, but I am not expert in these matters. Are you looking for a bijective proof, or just any old proof?

Comment Source:You could ask on MathOverflow. Or, if you'd rather not, someone could ask on your behalf. I expect that such a hypergeometric identity would succumb to Wilf-Zeilberger, but I am not expert in these matters. Are you looking for a bijective proof, or just any old proof?
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2.

I agree about Wilf-Zeilberger, and have been reading A=B. I don't have Maple or Mathematica, which may be necessary to solve this in a reasonable amount of time. I expect a question to MathOverflow would basically be asking someone to use a program for me. I'd prefer to understand something, and a bijective proof might help with that.

Comment Source:I agree about Wilf-Zeilberger, and have been reading A=B. I don't have Maple or Mathematica, which may be necessary to solve this in a reasonable amount of time. I expect a question to MathOverflow would basically be asking someone to use a program for me. I'd prefer to understand something, and a bijective proof might help with that.
• Options
3.

In reduce, I typed

load zeilberg;
summand:=k*(k+1)*(2*k+1)*factorial(n)*factorial(n-1)/(factorial(n-k)*factorial(n+k+1));
gosper(summand,k,1,n);


which resulted in $$\frac{(n+1)(n-1)!n}{(n+1)!}$$ which is a funny way of writing 1. I don't feel much wiser...

Comment Source:In [reduce](http://sourceforge.net/projects/reduce-algebra/), I typed ~~~~ load zeilberg; summand:=k*(k+1)*(2*k+1)*factorial(n)*factorial(n-1)/(factorial(n-k)*factorial(n+k+1)); gosper(summand,k,1,n); ~~~~ which resulted in $$\frac{(n+1)(n-1)!n}{(n+1)!}$$ which is a funny way of writing 1. I don't feel much wiser...