Generalizing the binomial theorem When $$(a+b)^n=\sum_{m=0}^{n}\mbox{}_nC_ma^{n-m}b^m$$ is generalized to any complex number $\alpha$ by $$(a+b)^{\alpha}=\sum_{n=0}^{\infty}\frac{\Gamma(\alpha+1)}{n!\Gamma(\alpha-n+1)},$$ is $n$ still restricted to integers? What if $\alpha=3.5$, does it mean that I will use $n=1,2,3$? I am confused that $n$ would go from 0 to infinity. 
Please help. Thank you.
 A: This is an application of Newtons binomial series
$$
(1+z)^α=\sum\binom{α}n z^n
$$
to $z=a/b$ or $z=b/a$. Please note the restriction $|z|<1$ for convergence. Also, the usual rules for products of powers and powers of products do not fully apply for complex $z$ and $α$.
See https://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalised_binomial_theorem
A: Let $ \displaystyle f(z) = (1+z)^{\alpha}$.
Then 
$$f'(z) = \alpha(1+ z)^{\alpha-1}$$
$$ f''(z) = \alpha(\alpha -1) (1+z)^{\alpha-2}$$
$$ f'''(z) = \alpha(\alpha-1)(\alpha-2) (1+z)^{\alpha-3}$$
$$ \text{etc}.$$
In general, if $ n >1$,
$$ f^{(n)}(x) = \alpha (\alpha-1) (\alpha-2) \cdots (\alpha - n+1) (1+z)^{\alpha-n}$$
The Taylor expansion of $f(z) = (1+z)^{\alpha}$ at $z=0$ is therefore
$$\begin{align} f(z)  &= \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} z^{n} \\ &= 1+ \sum_{n=1}^{\infty} \frac{\alpha(\alpha-1)(\alpha-2) \cdots (\alpha-n+1)}{n!} z^{n} \\ &= \sum_{n=0}^{\infty} \frac{\Gamma(\alpha +1)}{\Gamma(\alpha-n+1) n!} z^{n} \end{align}$$
which has a radius of convergence of $1$ since $f(z)$ has a branch point at $z=-1$.
So if $ \displaystyle |z| = \Big|\frac{b}{a}\Big| < 1$,
$$ f \left(\frac{b}{a}\right) = \frac{1}{a^{\alpha}} (a+b)^{\alpha} = \sum_{n=0}^{\infty}\frac{\Gamma(\alpha +1)}{\Gamma(\alpha-n+1) n!} \left(\frac{b}{a} \right)^{n}$$
which implies
$$(a+b)^{\alpha} = \sum_{n=0}^{\infty}\frac{\Gamma(\alpha +1)}{\Gamma(\alpha-n+1) n!} a^{\alpha-n} b^{n}$$
