Re: not so easy

glenn a écrit :
O/H Raymond Manzoni ??????:

Gamma(n/2+1/2)/Gamma(n/2+1) > Gamma(n/2+1)/Gamma(n/2+3/2)
is trivial since log(Gamma(x)) is convex nearly 'by definition' : http://en.wikipedia.org/wiki/Bohr-Mollerup_theorem so that
log(Gamma(n/2+1)) <= (log(Gamma(n/2+1/2))+log(Gamma(n/2+3/2)))/2
Gamma(n/2+1)^2 <= Gamma(n/2+1/2)*Gamma(n/2+3/2)
and Gamma(n/2+1/2)/Gamma(n/2+1) >= Gamma(n/2+1)/Gamma(n/2+3/2)

The geometric mean value of these values is sqrt(2/(n+1)) and must be between these values so that for all n>0 we have :
Gamma(n/2+1/2)/Gamma(n/2+1)>sqrt(2/(n+1))>Gamma(n/2+1)/Gamma(n/2+3/2)

Take the righ part and replace n+1 by n to get
(2/n)^(1/2) > Gamma(n/2+1/2)/Gamma(n/2+1) as wished

Hoping it clarified things!
Raymond

Now I've got it. So I guess that my friend either _proved_ in his mail the Bohr–Mollerup theorem, either he didn't new it so he tried with a different method.


Well I got the answer first using the Stirling series up to the 1/(12n) term but I had to separate the case n even (Gamma(n/2+1/2)/Gamma(n/2+1) = binomial(n, n/2) Gamma(1/2)/2^n) and n odd (2^n/[n binomial(n-1, (n-1)/2) Gamma(1/2)]) and take care of the error terms.
Doing this to the satisfaction of an examiner should require nearly the two pages!
Raymond
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