I need to prove:
$(1+z)^{\alpha }=\sum_{k=0}^{\infty }\binom{\alpha }{k}z^{k}$ for $|z|$<1 and $\alpha$ in
1) $\mathbb{N}$
2) $\mathbb{Z}$
I managed to prove it for natural $\alpha$ (by induction) but don't know how to do it for $\alpha$ in $\mathbb{Z}$. I don't know how to cope with negative binomial coefficients.
Furthermore I need to show that the equation above is true for $x $ in $ \mathbb{R}: |x| <1 $ and $\alpha=1/p$ (p in $\mathbb{N})$ and also for $\alpha$ in $\mathbb{Q}$
I'm thankful for every hint or push in the right direction you can give me!
For $\alpha=0$ it's clear. Now suppose that $\alpha\in -\Bbb N$. First note that: $${1\over{1+z}}=\Sigma_{m=0}^{\infty}(-1)^mz^m$$ then by differentiating we have: $${d^n{1\over{1+z}}\over{dz^n}}={{(-1)^nn!}\over{(1+z)^{n+1}}}=\Sigma_{m=n}^{\infty}n!{\binom mn}(-1)^mz^{m-n}$$ since the first $n$ terms get zero after differentiation. Therefore we obtain: $${{1}\over{(1+z)^{n+1}}}=\Sigma_{m=n}^{\infty}{\binom mn}(-1)^{m-n}z^{m-n}=\Sigma_{m=0}^{\infty}{\binom {m+n}n}(-1)^{m}z^{m}$$ by substituting $\alpha=-n-1$ we finally get $$(1+z)^{\alpha}=\Sigma_{m=0}^{\infty}{\binom {m-\alpha -1}{-\alpha -1}}(-1)^{m}z^{m}=\Sigma_{m=0}^{\infty}{\binom {\alpha}{m}}z^{m}$$