We have proved already (§ 215) that the sum of the series $1 + \binom{m}{1} z + \binom{m}{2} z^{2} + \dots$ is $$(1 + z)^{m} = \exp\{m\log(1 + z)\}$$, for all real values of $$m$$ and all real values of $$z$$ between $$-1$$ and $$1$$. If $$a_{n}$$ is the coefficient of $$z^{n}$$ then $\left|\frac{a_{n+1}}{a_{n}}\right| = \left|\frac{m – n}{n + 1}\right| \to 1,$ whether $$m$$ is real or complex. Hence (Ex. LXXX. 3) the series is always convergent if the modulus of $$z$$ is less than unity, and we shall now prove that its sum is still $$\exp\{m\log(1 + z)\}$$, the principal value of $$(1 + z)^{m}$$.

It follows from § 236 that if $$t$$ is real then $\frac{d}{dt}(1 + tz)^{m} = mz(1 + tz)^{m-1},$ $$z$$ and $$m$$ having any real or complex values and each side having its principal value. Hence, if $$\phi(t) = (1 + tz)^{m}$$, we have $\phi^{(n)}(t) = m(m – 1) \dots (m – n + 1)z^{n} (1 + tz)^{m-n}.$ This formula still holds if $$t = 0$$, so that $\frac{\phi^{n}(0)}{n!} = \binom{m}{n} z^{n}.$

Now, in virtue of the remark made at the end of § 164, we have $\phi(1) = \phi(0) + \phi'(0) + \frac{\phi”(0)}{2!} + \dots + \frac{\phi^{(n-1)}(0)}{(n – 1)!} + R_{n},$ where $R_{n} = \frac{1}{(n – 1)!}\int_{0}^{1} (1 – t)^{n-1} \phi^{(n)}(t)\, dt.$ But if $$z = r(\cos\theta + i\sin\theta)$$ then $|1 + tz| = \sqrt{1 + 2tr\cos\theta + t^{2}r^{2}} \geq 1 – tr,$ and therefore \begin{aligned} |R_{n}| &< \frac{|m(m – 1) \dots (m – n + 1)|}{(n – 1)!}\, r^{n} \int_{0}^{1} \frac{(1 – t)^{n-1}}{(1 – tr)^{n-m}}\, dt\\ &< \frac{|m(m – 1) \dots (m – n + 1)|}{(n – 1)!}\, \frac{(1 – \theta)^{n-1} r^{n}}{(1 – \theta r)^{n-m}},\end{aligned} where $$0 < \theta < 1$$; so that (cf. § 163) $|R_{n}| < K\frac{|m(m – 1) \dots (m – n + 1)|}{(n – 1)!}\, r^{n} = \rho_{n},$ say. But $\frac{\rho_{n+1}}{\rho_{n}} = \frac{|m – n|}{n}r \to r,$ and so (Ex. XXVII. 6) $$\rho_{n} \to 0$$, and therefore $$R_{n} \to 0$$, as $$n \to \infty$$. Hence we arrive at the following theorem.

Theorem. The sum of the binomial series $1 + \binom{m}{1} z + \binom{m}{2} z^{2} + \dots$ is $$\exp\{m\log(1 + z)\}$$, where the logarithm has its principal value, for all values of $$m$$, real or complex, and all values of $$z$$ such that $${|z|} < 1$$.

A more complete discussion of the binomial series, taking account of the more difficult case in which $$|z| = 1$$, will be found on pp. 225 et seq. of Bromwich’s Infinite Series.

Example XCVIII
1. Suppose $$m$$ real. Then since $\log(1 + z) = \tfrac{1}{2} \log(1 + 2r\cos\theta + r^{2}) + i\arctan\left(\frac{r\sin\theta}{1 + r\cos\theta}\right),$ we obtain \begin{aligned} \sum_{0}^{\infty} \binom{m}{n} z^{n} &= \exp\{\tfrac{1}{2}m \log(1 + 2r\cos\theta + r^{2})\} \operatorname{Cis} \left\{m\arctan\left(\frac{r\sin\theta}{1 + r\cos\theta}\right)\right\} \\ &= (1 + 2r\cos\theta + r^{2})^{\frac{1}{2}m} \operatorname{Cis} \left\{m\arctan\left(\frac{r\sin\theta}{1 + r\cos\theta}\right)\right\},\end{aligned} all the inverse tangents lying between $$-\frac{1}{2}\pi$$ and $$\frac{1}{2}\pi$$. In particular, if we suppose $$\theta = \frac{1}{2}\pi$$, $$z = ir$$, and equate the real and imaginary parts, we obtain \begin{aligned} 1 – \binom{m}{2} r^{2} + \binom{m}{4} r^{4} – \dots &= (1 + r^{2})^{\frac{1}{2}m} \cos(m\arctan r), \\ \binom{m}{1} r – \binom{m}{3} r^{3} + \binom{m}{5} r^{5} – \dots &= (1 + r^{2})^{\frac{1}{2}m} \sin(m\arctan r).\end{aligned}

2. Verify the formulae of Ex. 1 when $$m = 1$$, $$2$$, $$3$$. [Of course when $$m$$ is a positive integer the series is finite.]

3. Prove that if $$0 \leq r < 1$$ then \begin{aligned} 1 – \frac{1\cdot3}{2\cdot4} r^{2} + \frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8} r^{4} – \dots &= \sqrt{\frac{\sqrt{1 + r^{2}} + 1}{2(1 + r^{2})}}, \\ \frac{1}{2} r – \frac{1\cdot3\cdot5}{2\cdot4\cdot6} r^{3} + \frac{1\cdot3\cdot5\cdot7\cdot9}{2\cdot4\cdot6\cdot8\cdot10} r^{5} – \dots &= \sqrt{\frac{\sqrt{1 + r^{2}} – 1}{2(1 + r^{2})}}.\end{aligned}

[Take $$m = -\frac{1}{2}$$ in the last two formulae of Ex. 1.]

4. Prove that if $$-\frac{1}{4}\pi < \theta < \frac{1}{4}\pi$$ then \begin{aligned} \cos m\theta &= \cos^{m} \theta \left\{1 – \binom{m}{2} \tan^{2} \theta + \binom{m}{4} \tan^{4} \theta – \dots\right\}, \\ \sin m\theta &= \cos^{m} \theta \left\{\binom{m}{1} \tan\theta – \binom{m}{3} \tan^{3} \theta + \dots\right\},\end{aligned} for all real values of $$m$$. [These results follow at once from the equations $\cos m\theta + i\sin m\theta = (\cos\theta + i\sin\theta )^{m} = \cos^{m} \theta(1 + i\tan\theta)^{m}.]$

5. We proved (Ex. LXXXI. 6), by direct multiplication of series, that $$f(m, z) = \sum\dbinom{m}{n} z^{n}$$, where $$|z| < 1$$, satisfies the functional equation $f(m, z) f(m’, z) = f(m + m’, z).$ Deduce, by an argument similar to that of § 216, and without assuming the general result above, that if $$m$$ is real and rational then $f(m, z) = \exp\{m\log(1 + z)\}.$

6. If $$z$$ and $$\mu$$ are real, and $$-1 < z < 1$$, then $\sum \binom{i\mu}{n} z^{n} = \cos\{\mu\log(1 + z)\} + i\sin\{\mu\log(1 + z)\}.$