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I need to show the following identity. Any lead will be helpful and appreciated. Thanks in advance.

$$\binom{n}{m} m \sum_{i=0}^{m-1} \frac{\binom{m-1}{i}(-1)^i}{n-m+1+i}=1$$ for fixed $n$ and $m$.

pku
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2 Answers2

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$$S=\sum_{i=0}^{m-1}\binom{m-1}{i}\frac{(-1)^i}{n-m+i+1}=\int_{0}^{1}\sum_{i=0}^{m-1}\binom{m-1}{i}(-1)^i x^{n-m+i}\,dx $$ hence by the binomial theorem $$ S = \int_{0}^{1}\frac{x^n}{1-x}\left(-1+\frac{1}{x}\right)^{m}\,dx $$ and by setting $x=\frac{1}{z}$, then $z=t+1$, we get: $$ S = \int_{1}^{+\infty}\frac{(z-1)^{m-1}}{z^{n+1}}\,dz = \color{red}{\int_{0}^{+\infty}\frac{t^{m-1}}{(1+t)^{n+1}}\,dt}. $$ The last integral is easy to compute by integration by parts, or by using Euler's beta function.
At last we have: $$ S = \frac{\Gamma(m)\Gamma(n-m+1)}{\Gamma(n+1)} = \frac{m!(n-m)!}{m\cdot n!} = \color{red}{\frac{1}{m\binom{n}{m}}}$$ as wanted.

Jack D'Aurizio
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    Thank you Jack ! Thanks for the help. – pku Sep 21 '16 at 17:49
  • @ArnabKoley: you're welcome (but an upvote woud be kind if this answer actually helped you :) ) – Jack D'Aurizio Sep 21 '16 at 18:09
  • I went for upvote. But it says voters should have less than 15 reputations. I am new to this forum. Even if I can't caste my vote, I must say you helped me a lot. Thank you once again :) :) – pku Sep 21 '16 at 18:41
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The following MSE link proves that

$$\sum_{k=0}^n \frac{1}{x+k} (-1)^k {n\choose k} = n! \times \prod_{q=0}^n \frac{1}{x+q}.$$

Put $n=m-1$ and $x=n-m+1$ to get

$$(m-1)! \times \prod_{q=0}^{m-1} \frac{1}{n-m+1+q} \\ = (m-1)! \times \frac{(n-m)!}{n!} = \frac{1}{m} \times {n\choose m}^{-1}.$$

Community
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Marko Riedel
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