binomial expansion formula proof, bases on Lagrange form of Taylor series remainder Another exercise from Bartle/Sherbert Introduction to Real Analysis book (this one is exercise 9.4.14):
Use the Lagrange form of the remainder to justify the general Binomial Expansion
$$(1+x)^{m}=\sum_{n=0}^{\infty}\binom{m}{n}x^{n}\quad \mathrm{for}\ 0\le x<1$$
Note: $m$ in an arbitrary real number.
My take: To prove the statement, one should show that Taylor series coefficients around $x=0$ are indeed $\frac{f^{(n)}(0)}{n!}=\binom{m}{n}$, that is rather obvious, and then also that the limit of the Lagrange form of the remainder:
$$R_{n}(x)=\binom{m}{n+1}(1+c)^{m-(n+1)}x^{n+1}$$
is $0$ when $n\rightarrow\infty$ (of course, it is necessary to show this only for $0\le x<1$ and $0\le c<1$).
I have trouble with this later part.  Here, $\frac{x}{1+c}<1$, thus $(\frac{x}{1+c})^{n+1}\rightarrow 0<$ when $n\rightarrow\infty$. Also, $(1+c)^{m}$ won't affect the limit if it's zero. But I'm not sure what to do about $\binom{m}{n+1}$ part...
 A: If $m\in\mathbb{N}$, there is nothing to do. For non-integer $m>0$, let $k\ge0$ be an integer such that $k<m<k+1$. Then $0<m-k<1$, $-1<m-k-1<0$, $-2<m-k-2<-1$ and $n-k-1<n-m<n-k$. Noting that
\begin{eqnarray*}
\left|\binom{m}{n+1}\right|&=&\left|\frac{m(m-1)\cdots(m-n)}{(n+1)!}\right|\\
&\le&\frac{|m(m-1)\cdots(m-k+1)(m-k)(m-k-1)(m-k-2)\cdots(m-n)|}{(n+1)!}\\
&=&\frac{m(m-1)\cdots(m-k+1)(m-k)|(m-k-1)(m-k-2)\cdots(m-n)|}{(n+1)!}\\
&\le& m(m-1)\cdots(m-k+1)(m-k)\frac{1\cdot 2\cdots (n-k-1)}{(n+1)!}\\
&\to 0
\end{eqnarray*}
as $n\to\infty$. Thus
\begin{eqnarray*}
|R_n(x)|&=&\left|\binom{m}{n+1}(1+c)^{m-(n+1)}x^{n+1}\right|\\
&\le&\left|\binom{m}{n+1}\right|(1+c)^m\left(\frac{1}{1+c}\right)^{n+1}\\
&\le&\left|\binom{m}{n+1}\right|(1+c)^m\\
&\to&0
\end{eqnarray*}
For $m<0$, the story is different. We do not have $\left|\binom{m}{n+1}\right|\to 0$ as $n\to\infty$. In fact, let $k\ge 0$ be an integer such that $k\le-m<k+1$. Then
\begin{eqnarray*}
\left|\binom{m}{n+1}\right|&=&\left|\frac{m(m-1)\cdots(m-n)}{(n+1)!}\right|\\
&=&\left|\frac{-m(-m+1)\cdots(-m+n)}{(n+1)!}\right|\\
&\ge&\frac{(-m)(k+1)\cdots(k+n)}{(n+1)!}\\
&=&\frac{(-m)(n+k)!}{(n+1)!k!}\\
&\to \infty
\end{eqnarray*}
as $n\to\infty$. Thus $\lim_{n\to\infty}R_n(x)$ diverges.
A: Let me supply the proof that the remainder $R_{n-1}(x)$ tends to $0$ in the case $m<0$.
We can assume $0<c<x<1$ and let $\alpha=\frac{x}{1+c}<1$. It then suffices to show that $\binom{m}{n}\alpha^n=\binom{m}{n}(\frac x {1+c})^n\to 0$. 
Choose $k\in \mathbb N$ such that $k+1\ge -m$. Then for $\beta=\frac 1{\alpha}$ and $n\ge k$, 
\begin{align*}
\left|\frac{m(m-1)\cdots(m-n+1)}{n!}\alpha^n\right|
\le&\frac{(k+1)(k+2)\cdots(k+n)}{n!}\alpha^n\\
=& \frac{1}{k!}\cdot\frac{(k+n)!}{n!}\alpha^n\\
=& \frac{1}{k!}\cdot\frac{(n+1)(n+2)\cdots (n+k)}{\beta^n}\\
\le& \frac{1}{k!}\cdot\frac{(2n)^k}{\beta^n}\\
\to&0
\end{align*}
as $n\to \infty$. The last line follows from $\lim_{x\to \infty}\frac{x^k}{\beta^x}=0$, which in turn follows from L'Hospital rule.
