Problem

(a) State the sum of the series $z + z^2 + \cdots + z^n$, for $z \ne 1$.

(b) By letting $z = \frac{1}{2}(\cos\theta + \mathrm{i}\sin\theta)$ use de Moivre’s theorem to deduce that

$$\sum_{m = 1}^{n} \bigg(\frac{1}{2}\bigg)^m \sin m\theta = \frac{ \big(\frac{1}{2}\big)^{n + 2}\sin n\theta - \big(\frac{1}{2}\big)^{n + 1}\sin (n + 1)\theta + \frac{1}{2}\sin\theta }{\frac{5}{4} - \cos\theta}.$$

(c) Use the result in (b) to find $\sum_{m = 1}^{n} \big(\frac{1}{2}\big)^m m\cos m\theta$ in terms of $n$ and $\theta$. [You do not need to simplify your answer.]

(d) Hence find $\sum_{m = 1}^{\infty} \big(\frac{1}{2}\big)^m m\cos m\theta$ in terms of $\cos\theta$. [You may assume that $\frac{n}{2^n} \to 0$ as $n \to \infty$.]