yungman
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Can you show me how to get the series representation of \Gamma(n-3/2+1)?
For example \Gamma(n+3/2+1)=\frac{(2n+3)(2n+1)!}{2^{2n+2}.n!}.
I cannot figure out how to write a series with:
n=0 => \Gamma(0-3/2+1)= -2\sqrt{\pi}
n=1 => \Gamma(1-3/2+1)= \sqrt{\pi}
n=2 => \Gamma(2-3/2+1)= 1/2\sqrt{\pi}
n=3 => \Gamma(3-3/2+1)= 4/3\sqrt{\pi}
This is not homework. I have spent 2 days on this and can't figure it out!
Thanks
Alan
For example \Gamma(n+3/2+1)=\frac{(2n+3)(2n+1)!}{2^{2n+2}.n!}.
I cannot figure out how to write a series with:
n=0 => \Gamma(0-3/2+1)= -2\sqrt{\pi}
n=1 => \Gamma(1-3/2+1)= \sqrt{\pi}
n=2 => \Gamma(2-3/2+1)= 1/2\sqrt{\pi}
n=3 => \Gamma(3-3/2+1)= 4/3\sqrt{\pi}
This is not homework. I have spent 2 days on this and can't figure it out!
Thanks
Alan