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I struggled quite a while without success, so I highly appreciate if anybody can help me, proving that: $$ \frac{1}{\left(1+1/n \right)^{(M)}}=\sum_{k=0}^M \frac{(-1)^k}{k!(M-k)!(nk+1)}, $$ where $(1+1/n)^{(M)}$ is the rising factorial, with $ x^{(n)}=\frac{\Gamma(x+n)}{\Gamma(x)}$.

A correct answer would help answering this question.

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up vote 1 down vote accepted

Since Antonio Vargas helped a lot in answering this question, I will just briefly give a summary.

First multiply with $M!$ and recognize the Beta Function (see e.g. equation (21) here)

$$ \begin{eqnarray*} \frac{M!}{\left(1+1/n \right)^{(M)}}&=&\frac{\Gamma(M+1)\Gamma(1+1/n)}{\Gamma(M+1+1/n)} \\ &=&\frac{2}{n} \int_0^1 x^{\frac{2}{n}-1} (1-x^2)^{M} \,dx\\ &=&\int_0^1 (1-u^n)^M \,du \,\,(\text{ substitute } x = u^{n/2})\\ &=&\sum_{k=0}^M(-1)^k \frac{ M!}{k!(M-k)!(nk+1)}\\ \end{eqnarray*} $$

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