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Show $\sum_{n=0}^{\infty}\binom{n+k}{k}z^{n}=\frac{1}{(1-z)^{k+1}}$ where $|z|<1$ and $k \geq 0$.

I know

The right hand side: \begin{align*} \frac{1}{(1-z)^{k+1}} & = \sum_{n=k}^{\infty}n(n-1)...(n-k+1)z^{n-k} \\ \\ & = \sum_{n'=0}^{\infty}(n'+k)(n'+k-1)\ldots (n'+1)z^{n'} \ \text{let} \ n' = n-k \end{align*}

The left hand side: $$\sum_{n=0}^{\infty}\binom{n+k}{k}z^{n}=\frac{(n+k)(n+k-1)\ldots (n+1)}{k!}z^{n}$$

There is a $k!$ in the denominator. I stuck here.

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    $\begingroup$ No, that's not the right hand side. How did you get that equality? $\endgroup$ – Thomas Andrews Feb 26 '15 at 19:42
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First note

$$\binom{n+k}{k} = \binom{n+k}{(n+k) - k} = \binom{n+k}{n}.$$

Now

\begin{align}\binom{n+k}{n} &= \frac{(k+1)(k+2)\cdots (n+k)}{n!}\\ &= (-1)^n\frac{(-k-1)(-k-2)\cdots (-n-k)}{n!}\\ &= (-1)^n \frac{(-k-1)((-k-1)-1)\cdots ((-k-1)-n+1))}{n!}\\ &= (-1)^n\binom{-k-1}{n}, \end{align}

and hence

$$\sum_{n = 0}^\infty \binom{n+k}{k}z^n = \sum_{n = 0}^\infty \binom{-k-1}{n}(-z)^n = (1 + z)^{-k-1}.$$

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