163. Application to the binomial series

If $f(x)=(1+x{)}^m$, where $m$ is not a positive integer, then Cauchy’s form of the remainder is $$R_n=\frac{m(m–1)\dots (m–n+1)}{1\cdot 2\dots (n–1)}\frac{(1–\theta {)}^{n-1}{x}^n}{(1+\theta x{)}^{n-m}}.$$

Now $(1–\theta )/(1+\theta x)$ is less than unity, so long as $-1<x<1$, whether $x$ is positive or negative; and $(1+\theta x{)}^{m-1}$ is less than a constant $K$ for all values of $n$, being in fact less than $(1+|x|{)}^{m-1}$ if $m>1$ and than $(1–|x|{)}^{m-1}$ if $m<1$ Hence $$|R_n|<K|m|\left|(\genfrac{}{}{0ex}{}{m–1}{n–1})\right||x^n|={\rho }_n,$$ say. But ${\rho }_n\to 0$ as $n\to \mathrm{\infty }$, by Ex. XXVII. 13, and so $R_n\to 0$. The truth of the Binomial Theorem is thus established for all rational values of $m$ and all values of $x$ between $-1$ and $1$. It will be remembered that the difficulty in using Lagrange’s form, in Ex. LVI. 2, arose in connection with negative values of $x$.