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In trying to write an answer to this question:

calculate the roots of $z = 1 + z^{1/2}$ using Lagrange expansion

I have come across the identity

$$ \frac{1}{4^n}\binom{1/2}{n} = \frac{1}{1+2n}\binom{n+1/2}{2n}. \tag{1} $$

Could anyone help me prove this? The few identities I know for binomial coefficients aren't enough to get anywhere useful and I don't see a way to account for the large difference in the number of factors in their respective numerators and denominators.

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Simply compute ratios of each side. Introducing notation $\Theta_n f(n) = \frac{f(n+1)}{f(n)}$: $$ \Theta_n \frac{1}{4^n} \binom{1/2}{n} = \frac{1}{4} \Theta_n \binom{1/2}{n} =\frac{1}{4} \frac{\Gamma(n+1) \Gamma\left(\frac{3}{2}-n\right)}{\Gamma(n+2) \Gamma\left(\frac{1}{2}-n\right)} = -\frac{2n-1}{8(n+1)} $$ On another hand: $$ \Theta_n \frac{1}{2n+1} \binom{n+1/2}{2n} = \frac{2n+1}{2n+3} \Theta_n \binom{n+1/2}{2n} = \frac{2n+1}{2n+3} \frac{\frac{\Gamma(n+5/2)}{\Gamma(2n+3) \Gamma(1/2-n)}}{\frac{\Gamma(n+3/2)}{\Gamma(2n+1) \Gamma(3/2-n)}} = \frac{2n+1}{2n+3} \frac{(n+3/2)(1/2-n)}{(2n+2)(2n+1)} = -\frac{1}{8} \frac{2n-1}{n+1} $$ The identity is obviously true for $n=0$, which finishes the proof.

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  • 2
    $\begingroup$ That is simple! Thank you. $\endgroup$ – Antonio Vargas May 7 '13 at 19:07

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