TITLE: Taylor expansion of $(1+x)^α$ to binomial series – why does the remainder term converge?
QUESTION [16 upvotes]: For $α ∈ ℝ$ the function $g_α \colon B_1(0) → ℝ, x ↦ (1+x)^α$ is $C^∞$ and $g_α^{(n)}(x) = n! \tbinom{α}{n}(1+x)^{α-n}$, where $\tbinom{α}{n} = \frac{α(α-1)\cdots(α-n+1)}{n!}$ is the generalized binomial coefficient.
I want to show that $g_α$ has a Taylor expansion $g_α (x) = \sum_{k=0}^∞ \tbinom{α}{k}x^k$. I can easily show that the series really converges for $|x| < 1$, but can I use this fact to show that the remainder term
$$R_n(x) = \int_0^x \frac{(x-t)^n}{n!} (n+1)! \tbinom{α}{n+1}(1+t)^{α-n-1} dt$$
or in Lagrange form
$$\quad R_n(x) = \frac{x^{n+1}}{(n+1)!} (n+1)! \binom{α}{n+1}(1+ξ)^{α-n-1} = x^{n+1} \binom{α}{n+1}(1+ξ)^{α-n-1}$$
converges to zero for $|x| < 1$?
How else can I show that the remainder term converges to zero for $|x| < 1$?

Progress: Using the Lagrange form, for $x ∈ (-1,1)$ and $n ∈ ℕ$ there is an $h=h(n,x) ∈ (0,1)$ such that:
$$|R_n(x)| = \Big| x^{n+1} \binom{α}{n+1}(1+hx)^{α-n-1} \Big| = \Big| \binom{α}{n+1}\big(\frac{x}{1+hx}\big)^{n+1}(1+ξ)^{α} \Big| $$
If $x > 0$, then $\big|\frac{x}{1+hx}\big| < {\big|\frac{x}{1+x}\big|}< 1$ and so for $q := \big|\frac{x}{1+x}\big|$, since $\sum_{k=0}^∞ \tbinom{α}{n} q^n$ converges:
$$|R_n(x)| ≤ \Big| \binom{α}{n+1} · q^{n+1} ·(1+ξ)^{α} \Big| \overset{n → ∞}{\longrightarrow} 0$$
The problem is still open for $x < 0$.
Mhenni suggested to use Stirling approximation, but I want to do this more elementary and I think it is possible.

REPLY [2 votes]: Let $f(z)=(1+z)^\alpha$ and $g(z)=\sum_{n\geq 0}\binom{\alpha}{n}z^n$.
If $\alpha$ is a nonnegative integer, it is clear that both functions coincide on $\mathbb{C}$ by the binomial theorem. So we assume that $\alpha\not\in\mathbb{N}$ from now on, so that $\binom{\alpha}{n}\neq 0$ for every $n\geq 0$.
Using the principal branch of the complex logarithm, we see that $g(z)=\exp(\alpha\log(1+z))$ is holomorphic on $\mathbb{C}\setminus (-\infty,-1]$, whence on the open unit disk $D$ in particular.
Now observe that 
$$
\frac{\Big|\binom{\alpha}{n+1}\Big||z|^{n+1}}{\Big|\binom{\alpha}{n}\Big||z|^{n}}=\frac{|\alpha-n||z|}{n+1}\longrightarrow |z|.
$$
So the radius of convergence of $g$ is $1$ by the ratio test. Hence $g$ converges to a holomorphic function on $D$.
For every $x>0$, the remainder in the approximation of $f$ by its degree $n$ Taylor polynomial at $0$ is, by Taylor-Lagrange,
$$
R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}=\binom{\alpha}{n+1}(1+c)^{\alpha-n-1}x^{n+1}=(1+c)^\alpha\binom{\alpha}{n+1}\left(\frac{x}{1+c} \right)^{n+1}
$$
for some $0<c<x$. Since $0<\frac{x}{1+c}<1$, the convergence of $g$ at $\frac{x}{1+c}$ implies that the rhs above tends to $0$, since it is a constant times its general term. So the remainder of the Taylor approximation tends to $0$. 
It follows that $f$ and $g$ coincide on $(0,1)$.
By the isolated zero principle, $f$ and $g$ coincide on the whole open unit disk $D$.