If $$f(x) = (1 + x)^{m}$$, where $$m$$ is not a positive integer, then Cauchy’s form of the remainder is $R_{n} = \frac{m(m – 1)\dots (m – n + 1)}{1\cdot2\dots (n – 1)}\, \frac{(1 – \theta )^{n-1} x^{n}}{(1 + \theta x)^{n-m}}.$

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 $|R_{n}| < K |m| \left|\binom{m – 1}{n – 1}\right| |x^{n}| = \rho_{n},$ say. But $$\rho_{n} \to 0$$ as $$n \to \infty$$, 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 $$x$$.