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Notice that the geometric series $$\sum_{n \geq 0}{x^n}=\frac{1}{1-x}, x \in (-1,1)$$ is a special case of the binomial theorem for real exponents, in which the binomial theorem for real exponenets is $$(1+x)^a=\sum_{n \geq 0}{a \choose n}{x^n}$$ . The special case is when $a=-1$ and $x=-x$. I have no idea how to prove that $${-1 \choose n}{(-1)^n}=1, n \in Z^+$$

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Do you know the definition of $n\choose k$ for general $n$ and $k$? – MJD Jan 4 '13 at 3:39
General $n$ and positive integer $k$ will do. – Gerry Myerson Jan 4 '13 at 3:40
up vote 3 down vote accepted

For general $ n \in \mathbb{Z} $ and $ k \in \mathbb{N} $, we define $$ \binom{n}{k} \stackrel{\text{def}}{=} \frac{n \times (n - 1) \times \cdots \times (n - k + 1)}{1 \times 2 \times \cdots \times k}. $$ If you use this definition, you will obtain what you want. :)

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One possible definition of $\dbinom{\alpha}{\beta}$ is $$\dbinom{\alpha}{\beta} = \dfrac{\Gamma(\alpha + 1)}{\Gamma(\beta + 1) \Gamma(\alpha - \beta)}$$ When $\alpha$ happens to be negative integers say $\alpha_{\mathbb{Z}}$ and $\beta$ happens to be an integer $\in (-\infty,\alpha_{\mathbb{Z}}] \cup [0,\infty)$, then define $\dbinom{\alpha_{\mathbb{Z}}}{\beta}$ as $$\dbinom{\alpha_{\mathbb{Z}}}{\beta} = \lim_{\alpha \to \alpha_{\mathbb{Z}}}\dfrac{\Gamma(\alpha + 1)}{\Gamma(\beta + 1) \Gamma(\alpha - \beta)}$$

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