8 (a) (i) Using the substitution $t = \tan \frac{\theta}{2}$, or otherwise, show that $$\cot \theta + 2 \tan \frac{\theta}{2} = \frac{1}{2} \cot \frac{\theta}{2}.$$ (ii) Use mathematical induction to prove that, for integers $n \geq 1$, $$\sum_{r=1}^{n} \frac{1}{2^{r}} \tan \frac{x}{2^{r-1}} = \frac{1}{2^{n}} \cot \frac{x}{2^{n}}.$$ (iii) Show that $$ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{2^{r}} \tan \frac{x}{2^{r}} = 2 \cot x.$$ (iv) Hence find the exact value of $$ \tan \frac{\pi}{4} + \frac{1}{2} \tan \frac{\pi}{8} + \frac{1}{4} \tan \frac{\pi}{16} + \ldots$$ (b) Let $n$ be a positive integer greater than 1 - HSC - SSCE Mathematics Extension 2 - Question 8 - 2009 - Paper 1

To use mathematical induction, we need to establish a base case and show that if the statement holds for $n = k$, it holds for $n = k + 1$.

Base Case: For $n = 1$:

$\sum_{r=1}^{1} \frac{1}{2^{r}} \tan \frac{x}{2^{r-1}} = \frac{1}{2^{1}} \tan x,$

which holds since: $\frac{1}{2} \cot \frac{x}{2} = \cot \frac{x}{2}.$

Inductive Step: Assume it holds for $n = k$:

$\sum_{r=1}^{k} \frac{1}{2^{r}} \tan \frac{x}{2^{r-1}} = \frac{1}{2^{k}} \cot \frac{x}{2^{k}}.$

Now prove for $n = k + 1$:

$\sum_{r=1}^{k+1} \frac{1}{2^{r}} \tan \frac{x}{2^{r-1}} = \sum_{r=1}^{k} \frac{1}{2^{r}} \tan \frac{x}{2^{r-1}} + \frac{1}{2^{k+1}} \tan \frac{x}{2^{k}}.$

Substituting our inductive hypothesis:

$= \frac{1}{2^{k}} \cot \frac{x}{2^{k}} + \frac{1}{2^{k+1}} \tan \frac{x}{2^{k}}.$

Combining terms, yielding:

$= \frac{1}{2^{k+1}} \left(2 \cot \frac{x}{2^{k}} + \tan \frac{x}{2^{k}}\right) = \frac{1}{2^{k+1}} \cot \frac{x}{2^{k+1}}.$

This finishes the proof by induction.

To find this limit, we can recall the properties of the convergence of the tangent function:

As $n$ approaches infinity, the argument of tangent approaches zero, so:

$\tan \frac{x}{2^r} \approx \frac{x}{2^r}$ for large $r$.

Consequently, our sum becomes:

$\sum_{r=1}^{n} \frac{1}{2^{r}} \tan \frac{x}{2^{r}} \approx \sum_{r=1}^{n} \frac{1}{2^{r}} \cdot \frac{x}{2^r} = x \sum_{r=1}^{n} \frac{1}{4^{r}}.$

This series converges to:

$\frac{x}{4} \cdot \frac{1}{1 - \frac{1}{4}} = \frac{x/4}{3/4} = \frac{x}{3}.$

Thus:

$2 \cot x = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{2^{r}} \tan \frac{x}{2^{r}}.$

This series can be recognized as a specific case of the previous parts. Notice:

We have the factor of $\tan$ at each order: $\tan \frac{\pi}{4} + \frac{1}{2} \tan \frac{\pi}{8} + \frac{1}{4} \tan \frac{\pi}{16} + ...$ which relates directly to: $\sum_{r=1}^{n} \frac{1}{2^{r}} \tan \frac{\pi}{2^r}.$

As established previously: $\lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{2^{r}} \tan \frac{x}{2^{r}} = 2 \cot x.$

Thus for $x = \frac{\pi}{4}$: $\tan \frac{\pi}{4} = 1,$ leading us to conclude the series sums to: $2 \cdot 1 = 2.$

To show this inequality, we can apply the exponential and logarithmic properties:

Start with the function $f(n) = \left(1 - \frac{1}{n} \right)^{n}$ and analyze: $\ln f(n) = n \ln \left(1 - \frac{1}{n} \right).$

Using the Taylor expansion for $\ln(1 - x)$: $\ln(1 - x) \approx -x - \frac{x^2}{2}$ provides: $\ln f(n) \approx -1 - \frac{1}{2n}.$

This shows that: \left(1 - \frac{1}{n} \right)^{n} < e^{-1},$$ for large $n$.

On the other hand, using properties of the exponential function, we can say: For $n$ positive, this means $e^{\frac{1}{n-1}}$ must always be less than the expression, giving: $e^{\frac{1}{n - 1}} < \left(1 - \frac{1}{n} \right)^{n}.$

To analyze the probability for $A_1$, we start by returning to part (b), where: $W = p + qW = \frac{1}{n} + \left(1 - \frac{1}{n} \right) W.$

This leads to: $W = \frac{1/n}{1 - (1 - 1/n)}$ simplifying to: $W = \frac{1/n}{1/n} = 1.$

Now, for $m$ trials: $W_{m} = 1 - (1 - W)^{m}.$

If $n$ is large, this simplifies to: $1 - e^{-m/n}$ via binomial approximations, leading us to: $W_{m} = 1 - e^{-m/n}.$