Spivak, Ch. 23, "Infinite Series", Problem 21c The following problem is from Chapter 23 "Infinite Series" of Spivak's Calculus. Note that there are questions about convergence of the binomial series such as this one, but the current question concerns a specific proof. Namely, the one presented in the problem below.



*In this problem we will establish the "binomial
series"

$$(1+x)^\alpha=\sum\limits_{k=0}^\infty \binom{\alpha}{k}x^k, |x|<1$$
for any $\alpha$, by showing that $R_{n,0}(x)=0$. The proof is in
several steps, and uses the Cauchy and Lagrange forms as found in
problem 10-21.
(a) Use the ratio test to show that the series
$\sum\limits_{k=0}^\infty \binom{\alpha}{k} r^k$ does indeed converge for  > $|r|<1$.
(b) Suppose first that $0\leq x<1$. Show that $\lim\limits_{n\to\infty}
 R_{n,0}(x)=0$, but using Lagrange's form of the remainder, noticing
that $(1+t)^{\alpha-n-1}\leq 1$ for $n+1>\alpha$.
(c) Now suppose that $-1<x<0$; the number $t$ in Cauchy's form of the
remainder satisfies $-1<x_t\leq 0$. Show that
$$|x(1+t)^{\alpha-1}|\leq |x|M, \text{ where }
 M=\max(1,(1+x)^{\alpha-1}),$$
and
$$\left | \frac{x-t}{1+t} \right |=|x|\left ( \frac{1-t/x}{1+t} \right
 )\leq |x|\tag{1}$$
Using Cauchy's form of the remainder, and the fact that
$$(n+1)\binom{\alpha}{n+1} = \alpha \binom{\alpha-1}{n}$$
show that $\lim\limits_{n\to\infty} R_{n,0}(x)=0$

I was able to prove (a) and (b) and most of (c), up to (1). The issue is proving the last part, namely that $\lim\limits_{n\to\infty} R_{n,0}(x)=0$.
The solution manual has the following

$$|R_{n,0}(0)|=\left | (n+1)\binom{\alpha}{n+1}x(1+t)^{\alpha-1} \left(\frac{x-t}{1+t}\right )^n \right |\tag{2}$$
$$\leq |x\alpha M|\cdot \left |\binom{\alpha-1}{n}x^n\right | \to 0
 \text{ by part } a$$

First off, the $R_{n,0}(0)$ seems to be a typo. This would be the n-th order remainder evaluated at $0$.
What I have is the following
$$R_{n,0}(x)=\binom{\alpha}{n+1}(1+t)^{\alpha-(n+1)}x^{n+1}$$
$$=\binom{\alpha}{n+1}\cdot x\cdot (1+t)^{\alpha-1}\cdot \left ( \frac{x}{1+t}\right )^n$$
How is (2) obtained?
 A: Let $f(x)=(1+x)^\alpha$, $\alpha\neq0$.
Part (a) of the exercise (presumably solved)  implies that $\sum_n\binom{\alpha}{n}x^n$ converges for all $|x|<1$. The implies  that $\lim_{n\rightarrow}\binom{\alpha}{n}x^n=0$ for all  each $x$ with $|x|<1$.
Notice that
$$f^{(n)}(x)=\alpha\cdot\ldots\cdot(\alpha-n+1)(1+x)^{\alpha-n}=n!\binom{\alpha}{n}(1+x)^{\alpha-n}$$

(c) The residue in the Taylor expansion of $f$ around $x_0=0$ in Cauchy's form is
$$R_n(x)=\frac{1}{(n+1)!}f^{(n+1)}(\xi)(x-\xi)^n x=\binom{\alpha}{n+1}(1+\xi)^{\alpha-n-1}(x-\xi)^nx$$
where $\xi=\xi_{x, n}$ is between $x$ and $0$.
If $-1<x<0$, then $-1<x<\xi<0$ and so
$$0<\frac{\xi-x}{1+\xi}<-x=|x|$$
since $\phi(\xi)=\frac{\xi-x}{1+\xi}=1-\frac{x+1}{1+\xi}$ increases on $(-1,\infty)$. Indeed, $\phi'(\xi)=\frac{1+x}{(1+\xi)^2}>0$; hence $\phi(\xi)<\phi(0)=-x$ for all $x<\xi<0$. Consequently,
$$|R_n(x)|<\Big|\binom{\alpha}{n+1}\Big||x|^{n+1}(1+\xi_{x,n})^{\alpha-1}\xrightarrow{n\rightarrow\infty}0$$
since $0<1+x<|1+\xi_{x,n}|<1$.

(b) The remainder in the Taylor expansion of $f$ around $x_0=0$ in Lagrange's form is
$$R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!} x^n=\binom{\alpha}{n+1}(1+\xi)^{\alpha-n-1}x^{n+1}$$
where $\xi=\xi_{x, n}$ is between  $0$ and  $x$.
Therefore, for $0<x<1$, $0<\xi<x$ and so,
$$|R_n(x)|\leq\Big|\binom{\alpha}{n+1}x^{n+1}\Big|\xrightarrow{n\rightarrow\infty}0$$

Although not part of the questions mentioned in the OP, it is interesting determine when  $\sum_n\Big|\binom{\alpha}{n}\Big|$ converges.
Notice that for $n>\lfloor \alpha\rfloor$
$$n\Big(\frac{|a_{n+1}|}{|a_n|}-1\Big)=n\Big(\frac{n-\alpha}{n+1}-1\Big)=\frac{n}{n+1}(-\alpha-1)\xrightarrow{n\rightarrow\infty}-\alpha-1$$
From Raabe's theorem, it follows that if $\alpha>0$, $-\alpha-1<-1$ and so the series $\sum_n\Big|\binom{\alpha}{n}\Big|<\infty$; whereas if $\alpha<0$, $-1-\alpha>-1$ in which case, $\sum_n\Big|\binom{\alpha}{n}\Big|=\infty$.
