# A combinatorial identity: $\sum_{k=m}^n \frac{\binom{1/2}{k-m}}{k \binom{-1/2}{k}}=\frac{\binom{-1/2}{n-m}}{m \binom{-1/2}{n}}$

Let $m,n$ be two positive integer, $n>m$. I have trouble proving that $$\sum_{k=m}^n \frac{\binom{1/2}{k-m}}{k \binom{-1/2}{k}}=\frac{\binom{-1/2}{n-m}}{m \binom{-1/2}{n}}$$ Any suggestions, please? Thank you very much.

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Use $\binom{x}{k}=\frac{x(x-1)\dots(x-k+1)}{k!}$ and simplify. – LutzL Jan 3 '14 at 19:35
The $m$ in the both denominators in the identity to be checked looks suspicious. Are you sure that you do not want $n$ in the denominator on the right? – Phira Jan 3 '14 at 19:39
Looks amusing. What's the source? – Grigory M Jan 3 '14 at 20:12
@Phira, I'm sure. – Mark Jan 3 '14 at 23:26
@GrigoryM, my teacher. – Mark Jan 3 '14 at 23:26

You have a summand that does not depend on $n$ and you have a conjecture for general $n$ (which is clearly true for $m=n$), so it just remains to check the induction step:
$$\frac{\binom{-1/2}{n-m}}{m \binom{-1/2}{n}}+\frac{\binom{1/2}{n-m+1}}{(n+1) \binom{-1/2}{n+1}}=\frac{\binom{-1/2}{n-m+1}}{m \binom{-1/2}{n+1}},$$