Problem source

Prove by induction, or otherwise, that, if $0<\theta<\pi$, 
\[
\frac{1}{2}\tan\frac{\theta}{2}+\frac{1}{2^{2}}\tan\frac{\theta}{2^{2}}+\cdots+\frac{1}{2^{n}}\tan\frac{\theta}{2^{n}}=\frac{1}{2^{n}}\cot\frac{\theta}{2^{n}}-\cot\theta.
\]
Deduce that 
\[
\sum_{r=1}^{\infty}\frac{1}{2^{r}}\tan\frac{\theta}{2^{r}}=\frac{1}{\theta}-\cot\theta.
\]

Solution source

Claim: $\displaystyle \sum_{r=1}^n \frac1{2^r} \tan \tfrac{\theta}{2^r} = \frac1{2^n}\cot \tfrac{\theta}{2^n} - \cot \theta$

Proof: (By Induction) Base case: $n = 1$

\begin{align*}
&& LHS &= \sum_{r=1}^1 \frac1{2^r} \tan \frac{\theta}{2^r} \\
&&&= \frac1{2} \tan \frac{\theta}{2}\\
\\
&& RHS &= \frac12 \cot \frac{\theta}{2} - \cot \theta \\
&&&= \frac12 \frac{1}{\tan \frac{\theta}{2}} - \frac{1-\tan^2 \frac{\theta}{2}}{2 \tan \frac{\theta}{2}} \\
&&&= \frac{1}{2} \tan \frac{\theta}{2} = LHS
\end{align*}

Therefore our base case is true.

Assume our statement is true for some $n=k$, then consider $n = k+1$, ie

\begin{align*}
\sum_{r=1}^{k+1} \frac1{2^r} \tan \tfrac{\theta}{2^r} &= \sum_{r=1}^{k} \frac1{2^r} \tan \tfrac{\theta}{2^r}  + \frac1{2^{k+1}} \tan \frac{\theta}{2^{k+1}} \\
&= \frac{1}{2^k} \cot \frac{\theta}{2^k} - \cot \theta + \frac{1}{2^{k+1}}\tan \frac{\theta}{2^{k+1}} \\
&= \frac{1}{2^{k+1}} \left (2 \cot \frac{\theta}{2^k} +\tan \frac{\theta}{2^{k+1}} \right) - \cot \theta \\
&= \frac{1}{2^{k+1}} \left (2\frac{1-\tan^2 \frac{\theta}{2^{k+1}}}{2 \tan \frac{\theta}{2^{k+1}}} + \tan \frac{\theta}{2^{k+1}} \right) - \cot \theta \\
&= \frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \\
\end{align*}

Therefore, since as $x \to 0, x\cot  x \to 1$ or $x \cot \theta x \to \frac{1}{\theta}$

\begin{align*}
\sum_{r=1}^{\infty}\frac{1}{2^{r}}\tan\frac{\theta}{2^{r}} &= \lim_{k\to \infty} \sum_{r=1}^{k}\frac{1}{2^{r}}\tan\frac{\theta}{2^{r}} \\
&= \lim_{k\to \infty} \left ( \frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \right) \\
&=  \lim_{k\to \infty}\frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta  \\
&=  \lim_{k\to \infty}\frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta  \\
&=  \lim_{k\to \infty}\frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta  \\
&= \frac{1}{\theta} - \cot \theta
\end{align*}