How to show Lagrange's remainder of the negative binomial expansion tends to zero?

Question

Giving a binomial $(1-x)^{-n}$, it is well known we can expand it like like: $$1 + \binom{n}{1}x + \binom{n+1}{2}x^2 +\ ...$$ I am trying to understand why this expansion converges with $(1-x)^{-n}$ for $|x|\leq1$. And for that I want to use Lagrange's remainder, so that it becomes (say for $x\in(0,1)$ and $\xi\in(0,1)$ $$lim_{m\to\infty}\frac{f^{(m)}(\xi)}{(m+1)!}x^{m+1} = lim_{m\to\infty}\frac{(n+m+1)!}{(n-1)!(m+1)!}\frac{1}{(1-\xi)^{n+m+1}}x^{m+1}$$ but this expression tends to infinity. Could you help me to understand is there a mistake and why we are eventually allowed to use this expansion?

Answer 1

There is a general problem in what you are trying to do: the number $\xi$ also depends on $m$. So if you cannot obtain a uniform bound, you cannot calculate the limit. Below I show a case where you can, when $x<1/2$.

One can calculate that $$ f^{(m)}(x)=\frac{(n+m-1)!}{(n-1)!}\,(1-x)^{-n-m} $$

We assume $0<\xi<x<1/2$. So $1-\xi>1-x$.

The Lagrange remainder is $$ \frac{f^{m+1}(\xi)}{(m+1)!}\,x^{m+1}=\frac{(n+m)!}{(n-1)!(m+1)!}\,\frac{x^{m+1}}{(1-\xi)^{n+m+1}}. $$ The parts that depend only on $n$ are irrelevant to the limit, so we may consider $$ \frac{(n+m)!}{(m+1)!}\,\frac{x^{m+1}}{(1-\xi)^{m+1}}. $$

From $x<1/2$, we get $1-\xi>1/2>x$, and we get $q=x/(1-\xi)<1$. So we have to take a limit of the form $$ \lim_{m\to\infty} \frac{(n+m)!}{(m+1)!}\,q^{m+1}. $$ We have $$ \frac{(n+m)!}{(m+1)!}=\prod_{k=2}^n(m+k)\leq (m+n)^n. $$ Then $$ \limsup_{m\to\infty} \frac{(n+m)!}{(m+1)!}\,q^{m+1}\leq \limsup_{m\to\infty} (m+n)^n\,q^{m+1}=0. $$ An easy way to see that the last limit is zero is to consider the quotients $$ \frac{(m+1+n)^n q^{m+1}}{(m+n)^nq^m}\xrightarrow[m\to\infty]{}q<1 $$