$(1+z)^{\alpha }=\sum_{k=0}^{\infty }\binom{\alpha }{k}z^{k}$ for $|z|<1$

Question

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) $\mathbb{Z}$

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 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.

I'm thankful for every hint or push in the right direction you can give me! Please not that I'm not allowed to use differtiation yet

Answer 1

Hint for the case ($\alpha\in\mathbb{Z}$)

At first, in general, for $x\in\mathbb{R}$ and $k\in\mathbb{N}$, we define: $$\binom{x}{k}=\frac{x(x-1)(x-2)\dots(x-k+1)}{k!}$$ Also, for every $n\in\mathbb{N}$ we have that: $$(1+x)^n=\sum_{k=0}^\infty\binom{n}{k}x^k=\sum_{k=0}^n\binom{n}{k}x^k$$ Note, that for natural numbers, this is a finite sum and the equality hold for every $x\in\mathbb{R}$, not only $x\in(-1,1)$ - this will be needed later. Now, for $\alpha=-1$, note that, for every $k\in\mathbb{N}$: $$\binom{-1}{k}=\frac{-1(-1-1)(-1-2)\dots(-1-k+1)}{k!}=\frac{(-1)^kk!}{k!}=(-1)^k$$ Also, note that - using geometric series: $$\begin{align*} (1+x)^{-1}&=\frac{1}{1+x}=\\ &=\frac{1}{1-(-x)}=\\ &=\sum_{k=0}^\infty(-x)^k\\ &=\sum_{k=0}^\infty(-1)^kx^k\\ &=\sum_{k=0}^\infty\binom{-1}{k}x^k \end{align*}$$ which is the requested.

Now, note also that every $\alpha\in\mathbb{Z}_-$ can be writen in the form $\alpha=-n$, for some $n\in\mathbb{N}$. So, in general, one has, for $\alpha\in\mathbb{Z}_-$: $$(1+x)^\alpha=\frac{1}{(1+x)^n}=\left(\frac{1}{1+x}\right)^n=\left(\frac{1+x-x}{1+x}\right)^n=\left(1-\frac{x}{1+x}\right)^n$$ So, from our result for natural numbers - since this result holds for every $x\in\mathbb{R}$, so for $-\frac{x}{1+x}$, with $|x|<1$, as well, we have that: $$\begin{align*} (1+x)^\alpha&=\left(1-\frac{x}{1+x}\right)^n\\ &=\sum_{k=0}^n\binom{n}{k}\left(-\frac{x}{1+x}\right)^k\\ &=\sum_{k=0}^n\binom{n}{k}(-1)^kx^k(1+x)^{-k}\\ \end{align*}$$ But now, this smells like strong induction. We have proved the case $n=1$ ($\alpha=-1$) and we se that in this recusrive relation, all the previous sums $(1+x)^{-k}$ for $k=0,1,\dots,n$ appear. Can you continue from this point? - You will need to prove some properties of $\binom{\alpha}{k}$ similar to the ones for natural numbers.