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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 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 at 19:39
    
Looks amusing. What's the source? –  Grigory M Jan 3 at 20:12
    
@Phira, I'm sure. –  Mark Jan 3 at 23:26
    
@GrigoryM, my teacher. –  Mark Jan 3 at 23:26

1 Answer 1

up vote 3 down vote accepted

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:

$$\frac{\binom{-1/2}{n-m}}{m \binom{-1/2}{n}}+\frac{\binom{1/2}{n-m+1}}{(n+1) \binom{-1/2}{n+1}}=\frac{\binom{-1/2}{n-m+1}}{m \binom{-1/2}{n+1}},$$

which reduces to a polynomial identity after you cancel all common factors.

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