Simplifying 'hypergeom'

Let us recall our starting point.

Int((1-t)/((t+1)*sqrt(t)),t = 0 .. 1) = 2/3*hypergeom([1, 2],[5/2],1/2);

Math

First we recall what 'hypergeom' stands for. It is the Gauss hypergeometric function (or series), and defined as

hypergeom([a, b],[c],z) = 'GAMMA(c)/(GAMMA(a)*GAMMA(b))* sum((GAMMA(a+n)*GAMMA(b+n)/GAMMA(c+n))*z^n/n!, n=0..infinity)';

Math

Because Math, Mathand Math, further Mathwe can simplify to

'(1/2)*sqrt(Pi)*sum((GAMMA(2+n)/GAMMA(5/2+n))*(1/2)^n,n=0..infinity)' =(1/2)*sqrt(Pi)*sum((GAMMA(2+n)/GAMMA(5/2+n))*(1/2)^n,n=0..infinity);

Math

Now if we set Mathand recall Mathwe arrive at

'(1/2)*sum((m-1)!/((2*m)!/(m!*4^m))*(1/2)^(m-2),m=2..infinity)' = (1/2)*sum((m-1)!/((2*m)!/(m!*4^m))*(1/2)^(m-2),m=2..infinity);

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This simplifies further:

'2*sum(2^m*GAMMA(m)^2*m/GAMMA(2*m+1),m=2..infinity)' = 2*sum(2^m*GAMMA(m)^2*m/GAMMA(2*m+1),m=2..infinity);

Math

'2*sum(2^m/(m*binomial(2*m,m)),m=2..infinity)' = 2*sum(2^m/(m*binomial(2*m,m)),m=2..infinity);

Math

'sum(2^(m+1)/(m*binomial(2*m,m)),m=1..infinity)' = sum(2^(m+1)/(m*binomial(2*m,m)),m=1..infinity);

Math

Another beautifull fomula derived from Euler's equation!