Generalised Binomial Theorem Intuition
Solution 1:
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 $|x|<1$.
Solution 2:
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 $|x|<1$ 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
Solution 3:
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.