Generalised Binomial Theorem Intuition It was not until recently (why don't they teach it in secondary school?) that I've come across the Generalised Binomial Theorem, which from what I can tell is basically the same as the regular Binomial Theorem, except that the finite sum is replace by an infinite series:
$$
(x+y)^n=\sum^{n}_{r=0}\binom{n}{r}x^{n-r}y^r=\sum^{\infty}_{r=0}\binom{n}{r}x^{n-r}y^r
$$
Unfortunately, I wasn't able to find any clear explanation of how to get from the regular theorem to the generalised one, the only proof I found being based on some obscure mathematics, while my math book entirely skips the explanation. 
Hence, my question is how do you prove the generalised theorem by deriving it from the regular theorem or otherwise?
 A: Here is another approach: The formula
$$(1+x)^\alpha=\sum_{k=0}^\infty {\alpha\choose k}\, x^k\qquad(1)$$
is true for $\alpha\in{\mathbb N}$; so maybe its true for arbitrary $\alpha$. To find out fix an $\alpha\in{\mathbb R}$ and consider the function
$$f(x):=\sum_{k=0}^\infty {\alpha\choose k}\, x^k\qquad(|x|<1)\ .$$
Using termwise differentiation one obtains
$$f'(x)=\sum_{k=1}^\infty {\alpha\choose k}\,k\, x^{k-1}=\sum_{k'=0}^\infty {\alpha\choose k'+1}\,(k'+1)\, x^{k'}=\sum_{k=0}^\infty {\alpha\choose k}\,(\alpha -k)\, x^k\ .$$
Therefore, using the third and the first  expression for $f'(x)$, we get
$$(1+x)f'(x)=\sum_{k=0}^\infty {\alpha\choose k}\,(\alpha -k)\, x^k+\sum_{k=0}^\infty {\alpha\choose k}\,k\, x^k=\alpha f(x)\ ,$$
or
$$(1+x)f'(x)-\alpha f(x)\equiv0\ .$$
It follows that ${d\over dx}\bigl((1+x)^{-\alpha} f(x)\bigr)\equiv0$, or $(1+x)^{-\alpha} f(x)={\rm const.}$, and as $f(0)=1$ one concludes that $f(x)\equiv (1+x)^\alpha$; whence $(1)$ is indeed true for $|x|<1$.
A: In addition (and for future reference) to everybody else's answers I have found the following to be true:
$$
\begin{align}
(x+y)^n&=\sum^{n}_{r=0}\binom{n}{r}x^{n-r}y^r &\qquad(1)\\
&=\sum^{\infty}_{r=0}\binom{n}{r}x^{n-r}y^r &\qquad(2)
\end{align}
$$


*

*Let $x,y\in\mathbb{C}\setminus\{0\}$. Eq. $(1)$ and $(2)$ are both true if and only if $n\in\mathbb N$.

*For $n\in\mathbb C$ and for $(2)$ to hold, the values of $x$ and $y$ must satisfy $|x|>|y|$.


A quick look at the case $y=1$ tells us that:
$$
\begin{align}
(1+x)^n&=\sum^{\infty}_{r=0}\binom{n}{r}x^r &\qquad(3)\\
&=\sum^{\infty}_{r=0}\binom{n}{r}x^{n-r} &\qquad(4)
\end{align}
$$


*

*When $n\in\mathbb{C}$ and $|x|&lt1$ we'd use $(3)$

*When $n\in\mathbb{C}$ and $|x|>1$ we'd use $(4)$

*When $n\in\mathbb{N}$ or $|x|=1$ either equation works

A: In case you don't know the Taylor-expansion yet:
1) For $n \in \mathbb{N}$ and $n &lt m$ we have $\binom{n}{m}$ = 0. Therefore further summation does not change the result.
2) For $n$ a fraction like -1/2 (and $y = 1, |x| &lt 1$) you can use the sum of the geometric series to get an infinite series
$$
\sqrt{\dfrac{1}{1+x}}\sqrt{\dfrac{1}{1+x}} = 1 - 1x^1 + 1x^2 +- ... = 
$$
To solve this, put
$$
\sqrt{\dfrac{1}{1+x}}= a_0 + a_1x^1 + a_2x^2 + ... 
$$
and multiply this with itself, ordering the result by exponents of $x$ and comparing with the coefficients 1 and -1, respectively, of the upper series.
A: Here is one straightforward approach, mentioned in Lang's Analysis book.
Let $ \alpha \in \mathbb{R} $ and $ x \in (-1,1) $.
We'll try to show $$ (1+x)^\alpha = 1 + \binom{\alpha}{1}x + \binom{\alpha}{2} x^2 + \ldots $$ holds (where for $ k \in \mathbb{Z}_{>0} $, $ \binom{\alpha}{k} := \frac{\alpha (\alpha - 1) \ldots (\alpha - (k-1))}{k!} $)
Writing $ (1+x)^\alpha = 1 + \binom{\alpha}{1} x + \binom{\alpha}{2} x^2 + \ldots + \binom{\alpha}{n-1} x^{n-1} + R_n (x) $, we now need to show $ R_n (x) \to 0 $ as $ n \to \infty $.

Recall integral version of Taylor's theorem

Let $ f : I \to \mathbb{R} $ be a $ C^n $ function on open interval $ I $, and say $ a, b \in I $. Then $$ f(b) = f(a) + f'(a) (b-a) + \ldots + \dfrac{f^{(n-1)}(a)}{(n-1)!} (b-a)^{n-1} + \int_{a}^{b} \dfrac{(b-t)^{n-1}}{(n-1)!} f^{(n)}(t) dt $$

So applying it to $ f : (-1, \infty) \to \mathbb{R} $ given by $ f(t) = (1+t)^\alpha $ and with $ a = 0 $,
$$ \begin{align} R_n (x) &= \int_{0}^{x} \dfrac{(x-t)^{n-1}}{(n-1)!} \alpha (\alpha - 1) \ldots (\alpha -(n-1)) (1+t)^{\alpha - n} dt \\ &= n \binom{\alpha}{n} \int_{0}^{x} (x-t)^{n-1} (1+t)^{\alpha - n} dt \end{align} $$

Estimating $\binom{\alpha}{n} $ term :
If $ \alpha = 0 $, $ \binom{\alpha}{n} $ is trivially $ 0 $. So lets take $ \alpha \neq 0 $.
$\begin{align} \left|\binom{\alpha}{n}\right| &= \left|\dfrac{\alpha (\alpha-1) \ldots (\alpha - (n-1))}{n!} \right| \\ &= \left| \dfrac{1}{n} \alpha \left( \dfrac{\alpha}{1} - 1 \right) \left( \dfrac{\alpha}{2} - 1 \right) \ldots \left( \dfrac{\alpha}{n-1} - 1 \right) \right| \\ &\leq \dfrac{1}{n} |\alpha| \left( \dfrac{|\alpha|}{1} + 1 \right) \left( \dfrac{|\alpha|}{2} + 1 \right) \ldots \left( \dfrac{|\alpha|}{n-1} + 1 \right) \\ \end{align}$
Since each factor in RHS is $ > 0 $, we can write it as
$$ e^{ \ln \left[ \frac{1}{n} |\alpha| \left( \frac{|\alpha|}{1} + 1 \right) \left( \frac{|\alpha|}{2} + 1 \right) \ldots \left( \frac{|\alpha|}{n-1} + 1 \right) \right] }. $$
The exponent here is
$\begin{align} &- \ln n + \ln|\alpha| + \ln\left(\frac{|\alpha|}{1} + 1\right) + \ln\left( \frac{|\alpha|}{2} + 1 \right) +   \ldots + \ln\left( \frac{|\alpha|}{n-1} + 1 \right) \\ &\leq - \ln n + \ln |\alpha | + \frac{|\alpha|}{1} + \frac{|\alpha|}{2} + \ldots + \frac{|\alpha|}{n-1} \end{align} $
[because $ \ln (1+x) \leq x $ for $ x \geq 0 $. This in turn is because $ x - \ln(1+x) $ is $ 0 $ at $ x = 0 $, and its derivative $ 1 - \frac{1}{1+x} $ is $ > 0 $ on $ (0, \infty) $]
$\begin{align} &\leq -\ln n + \ln |\alpha| + |\alpha| \left( 1 + \ln (n-1) \right) \end{align} $
[because $ 1 + \frac{1}{2} + \ldots + \frac{1}{n-1} $ $= 1 + \int_{1}^{2} \frac{1}{2} dt + \ldots + \int_{n-2}^{n-1} \frac{1}{n-1} dt $ $ \leq 1 + \int_{1}^{2} \frac{1}{t} dt + \ldots + \int_{n-2}^{n-1} \frac{1}{t} dt $ $ = 1 + \ln (n-1) $. This is also clear visually].
So finally,
$\begin{align} \left| \binom{\alpha}{n} \right| &\leq e^{ \ln \left[ \frac{1}{n} |\alpha| \left( \frac{|\alpha|}{1} + 1 \right) \left( \frac{|\alpha|}{2} + 1 \right) \ldots \left( \frac{|\alpha|}{n-1} + 1 \right) \right] } \\ &\leq e^{-\ln n + \ln |\alpha| + |\alpha| (1 + \ln (n-1) ) } \\ &= \frac{1}{n} |\alpha| e^{|\alpha|} (n-1)^{|\alpha|} ( \leq |\alpha| e^{|\alpha|} n^{|\alpha| - 1} ) \end{align} $
Especially,
$$ \fbox{$\left|\binom{\alpha}{n}\right| = \mathcal{O} (n^{|\alpha|-1}) $ } $$

Estimating $R_n(x)$ :
$ \underline{\text{CASE-1}} \, (x \in [0,1)) $
In the integral as $ t $ varies from $ 0 $ to $ x $, $ (1+t)^{\alpha - n} \leq (1+t)^{\alpha} \leq 2^\alpha $, so
$\begin{align} |R_n (x)| &= n \left| \binom{\alpha}{n}\right| \int_{0}^{x} (x-t)^{n-1}(1+t)^{\alpha - n} dt \\ &\leq n \left| \binom{\alpha}{n} \right| 2^\alpha \int_{0}^{x} (x-t)^{n-1} dt \\ &= \left| \binom{\alpha}{n} \right| 2^\alpha x^n   \end{align}$
Now RHS is $ \leq K n^{|\alpha|-1} x^n $, and $ n^{|\alpha|-1} x^n $ goes to $ 0 $ as $ n \to \infty $ [if $ x = 0 $, trivial. Else write $ x $ as $ \frac{1}{1+y} $ with $ y > 0 $ and expand the denominator], as needed.
$\underline{\text{CASE-2}} \, (x \in (-1, 0)) $
$\begin{align} R_n (x) &= n \binom{\alpha}{n} \int_{0}^{x} (x-t)^{n-1} (1+t)^{\alpha - n} dt \\ &= n \binom{\alpha}{n} \int_{0}^{-|x|} (-|x| - t)^{n-1} (1+t)^{\alpha - n} dt \end{align}$
Substituting $ s = (-t) $, the integral becomes
$\begin{align} R_n(x) &= n \binom{\alpha}{n} \int_{0}^{|x|} (-|x| + s)^{n-1} (1-s)^{\alpha - n} (-ds) \\ &= n \binom{\alpha}{n} (-1)^n \int_{0}^{|x|} (|x|-s)^{n-1} (1-s)^{\alpha - n} ds \end{align}$
and now keeping in mind $ |x| \in (0,1) $ and that $ s $ varies from $ 0 $ to $ |x| $ in the integral,
$\begin{align} |R_n (x)| &= n \left| \binom{\alpha}{n} \right| \int_{0}^{|x|} \left(\dfrac{|x|-s}{1-s} \right)^{n-1} (1-s)^{\alpha - 1} ds \\ &\leq n \left| \binom{\alpha}{n} \right| \int_{0}^{|x|} |x|^{n-1} (1-s)^{\alpha -1}ds \\ &\leq L n^{|\alpha|} |x|^{n-1}\end{align}$
and $ n^{|\alpha|} |x|^{n-1} $ goes to $ 0 $ as $ n \to \infty $, as needed.
