Answer: (i) For every positive integer \(n\),

\(1+z+z^2+\cdots+z^{n-1}=\dfrac{z^n-1}{z-1}\) for \(z eq 1\).

(ii) Hence

\(\displaystyle \sum_{m=1}^{\infty}\left(\frac12\right)^m\sin m\theta=\frac{2\sin\theta}{5-4\cos\theta}.\)

(i) We prove the result by induction on \(n\).

Let \(P(n)\) be the statement

\(1+z+z^2+\cdots+z^{n-1}=\dfrac{z^n-1}{z-1}\), where \(z eq 1\).

Base case: When \(n=1\), the left-hand side is just \(1\). The right-hand side is

\(\dfrac{z^1-1}{z-1}=\dfrac{z-1}{z-1}=1\).

So \(P(1)\) is true.

Inductive step: Assume \(P(k)\) is true for some positive integer \(k\), so

\(1+z+z^2+\cdots+z^{k-1}=\dfrac{z^k-1}{z-1}.\)

Then for \(k+1\),

\(1+z+z^2+\cdots+z^{k-1}+z^k = \dfrac{z^k-1}{z-1}+z^k.\)

Putting this over a common denominator gives

\(\dfrac{z^k-1}{z-1}+z^k = \dfrac{z^k-1+z^k(z-1)}{z-1} = \dfrac{z^{k+1}-1}{z-1}.\)

Thus \(P(k+1)\) is true whenever \(P(k)\) is true.

Therefore, by mathematical induction,

\(1+z+z^2+\cdots+z^{n-1}=\dfrac{z^n-1}{z-1}\)

for all positive integers \(n\), provided \(z eq 1\).

(ii) Let

\(z=\dfrac12(\cos\theta+i\sin\theta)=\dfrac12 e^{i\theta}.\)

Since \(|z|=\frac12<1\), we can use the geometric-series result from part (i) in the limit as \(n\to\infty\):

\(\sum_{m=0}^{\infty} z^m = \dfrac{1}{1-z}.\)

Therefore

\(\sum_{m=1}^{\infty} z^m = \dfrac{z}{1-z}.\)

Now take imaginary parts. Since \(z^m=\left(\frac12\right)^m(\cos m\theta+i\sin m\theta)\), the imaginary part of \(\sum_{m=1}^{\infty} z^m\) is exactly \(\sum_{m=1}^{\infty}\left(\frac12\right)^m\sin m\theta\).

So we compute

\(\sum_{m=1}^{\infty} z^m = \frac{z}{1-z} = \frac{\frac12(\cos\theta+i\sin\theta)}{1-\frac12(\cos\theta+i\sin\theta)}.\)

Multiply top and bottom by \(2\):

\(\frac{\cos\theta+i\sin\theta}{2-\cos\theta-i\sin\theta}.\)

Now multiply numerator and denominator by the conjugate \(2-\cos\theta+i\sin\theta\):

\(\frac{(\cos\theta+i\sin\theta)(2-\cos\theta+i\sin\theta)}{(2-\cos\theta)^2+\sin^2\theta}.\)

The denominator simplifies to

\((2-\cos\theta)^2+\sin^2\theta = 4-4\cos\theta+\cos^2\theta+\sin^2\theta = 5-4\cos\theta.\)

The numerator expands to

\(\cos\theta(2-\cos\theta)-\sin^2\theta + i\big(\sin\theta(2-\cos\theta)+\cos\theta\sin\theta\big).\)

The imaginary part is

\(2\sin\theta.\)

Hence

\(\sum_{m=1}^{\infty}\left(\frac12\right)^m\sin m\theta = \frac{2\sin\theta}{5-4\cos\theta}.\)