Recently, a student asked this summation $$ S= \sum_{k=0}^n \frac{{n \choose k}}{ {m + n-1 \choose k}} $$ without putting any effort. Interestingly, opening of the binomial coefficient doesn't seem to help, yet it is doable.

Use $$ {N\choose k}^{-1}=(N+1)\int_{0}^{1} x^k(1-x)^{N-k}\, dx. $$ Then $$ S= \sum_{k=0}^n \frac{{n \choose k}}{ {m + n-1 \choose k}}= (m+n) \sum_{k=0}^{n}\int_{0}^{1} {n \choose k} x^k (1-x)^{m+n-1}\, dx $$ $$\implies S=(m+n)\int_{0}^{1}\, dx~ (1-x)^{m+n-1}\sum_{k=0}^{n} {n \choose k}\left(\frac{x}{1-x}\right)^k$$ $$\implies S=(m+n)\int_{0}^{1}(1-x)^{m-1}dx=\frac{m+n}{m}.$$

The question is: How to do it otherwise?


Opening of the binomial coefficient indeed works. $$ \sum_{k=0}^n \frac{{n \choose k}}{ {m + n-1 \choose k}}\\=\frac{n!}{(m+n-1)!}\sum_{k=0}^n \frac{(m+n-k-1)!}{(n-k)!}\\=\frac{n!(m-1)!}{(m+n-1)!}\sum_{k=0}^n \frac{(m+n-k-1)!}{(n-k)!(m-1)!}\\=\frac{n!(m-1)!}{(m+n-1)!}\sum_{k=0}^n {m+n-k-1 \choose m-1}\\=\frac{n!(m-1)!}{(m+n-1)!}\sum_{k=0}^n {m+k-1 \choose m-1} $$
We know that, $$\sum_{k=0}^n {m+k-1 \choose m-1}={m-1 \choose m-1}+{m \choose m-1}+\cdots +{m+n-1 \choose m-1}={m+n \choose m}$$ Therefore, $$\sum_{k=0}^n \frac{{n \choose k}}{ {m + n-1 \choose k}}=\frac{n!(m-1)!}{(m+n-1)!}\cdot{m+n \choose m}=\frac{m+n}m$$

  • 2
    $\begingroup$ Great answer, in the ffourt step invoking $k \to n-k$ would have been a little better. $\endgroup$
    – Z Ahmed
    Jun 18 '21 at 3:51

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