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The complex power series $$\sum_{n=0}^\infty \dbinom{n+m}{m} z^n$$ has radius of convergence by Ratio Test also it converges to $\frac{1}{(1-z)^{m+1}}$ in $D(0,1)$. I proved this by noting $\frac{1}{1-z} = \sum z^n$ and that power series have infinite derivatives in their raidus of convergence, with their derivatives equal to the term-wise derivatives sum.

I was wondering if there is a proof from first principles without using infintie sum differentiation.

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  • $\begingroup$ You could take the $(m+1)$-fold Cauchy product of the geometric series. $\endgroup$ – Lord Shark the Unknown May 13 '17 at 18:20
  • $\begingroup$ Try showing that $\sum_{n=0}^\infty {n+m \choose m} z^n \sum_{k=0}^{m+1} {m+1 \choose k} z^k = 1$ $\endgroup$ – reuns May 13 '17 at 21:06
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There is in fact a simple identity of binomial coefficients that shows this: \begin{align} \binom{n+m}{m} &= \frac{(n+m)!}{m!n!} \\ &= \frac{(n+m)(n+m-1)\dotsb (m+2)(m+1)}{n!} \\ &= (-1)^n \frac{(-m-1)(-m-2)\dotsb ((-m-1)-n)((-m-1)-n+1)}{n!} \\ &= (-1)^n \binom{-m-1}{n} \end{align} (equally, the reflection identity implies that $\binom{n+m}{n} = \binom{n+m}{m} = (-1)^n \binom{-m-1}{n} $). Then one simply has $$ \sum_{n=0}^{\infty} \binom{n+m}{m} z^n = \sum_{n=0}^{\infty} (-1)^n \binom{-m-1}{n} z^n = \frac{1}{(1-z)^{m+1}} $$ by the generalised binomial theorem.

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  • $\begingroup$ Yup. I've written this in the answer to many questions here. $\endgroup$ – marty cohen May 14 '17 at 2:22

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