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I want to prove that $$ \binom nm \sum_{i=0}^{m} (-1)^i\frac{\binom mi}{n-m+1+i}=\frac{1}{1+n}\\ $$ $$ \binom nm [ \frac{\binom m0}{n-m+1} +(-1)^1\frac{\binom m1}{n-m+2}+... ..+(-1)^m\frac{\binom mm}{n+1}] =\frac{1}{1+n}\\ $$

  1. Assume that $m=1$,then it's equal to $1/(n+1)$. $$ \binom n1 [ \frac{1}{n} -\frac{1}{n+1}] =\frac{1}{1+n}\\ $$

  2. Assume that $m=2$,then it's equal to $1/(n+1)$.

$$ \binom n2 [ \frac{1}{n-1} -\frac{2}{n}+\frac{1}{n+1}] =\frac{1}{1+n}\\ $$

How to prove that for $\forall m$, this formula is equal to $1/(n+1)$？

ahuigo
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2 Answers2

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Hint. Note that $$x^{n-m}(1-x)^m=\sum_{i=0}^{m} (-1)^i \binom mi x^{n-m+i}$$ Now integrate with respect to $x$ over the interval $[0,1]$ and recall the definition of Beta function.

Robert Z
  • 145,942
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The partial fraction expansion of $$\frac1{x(x+1)(x+2)\cdots(x+n)}$$ is $$\frac1{m!}\sum_{i=0}^m(-1)^i\frac{\binom mi}{x+i}.$$ Now take $x=n-m+1$.

Angina Seng
  • 158,341
  • https://math.stackexchange.com/questions/715706/partial-fraction-expansion-of-frac1xx1x2-cdotsxn/715718 – ahuigo Jan 01 '19 at 11:07
  • https://math.stackexchange.com/questions/38623/how-to-prove-sum-limits-r-0n-frac-1rr1-binomnr-frac1n1 – ahuigo Jan 01 '19 at 11:09