## Answers to exercises

##### Exercise 1
$$2,\ 5,\ 26,\ 677,\ 458\,330$$.
##### Exercise 2
1. $$a_n = 2n$$
2. $$a_n = n^2$$.
##### Exercise 3

The sequence simplifies to

$\log_5 2,\ \ 2\log_5 2,\ \ 3\log_5 2,\ \dots$

and so the general term is $$a_n = n\log_5 2$$.

##### Exercise 4

Here $$a = 210$$ and $$d = -13$$, so the general term is given by $$a_n = 210 - 13(n-1) = 223-13n$$. The equation $$223 - 13n = 12$$ has solution $$n = \dfrac{211}{13}$$, which is not positive integer. Hence 12 is not a term in the sequence.

##### Exercise 5

We have $$a = \sqrt{6}$$ and $$r = \dfrac{2\sqrt{3}}{\sqrt{6}} = \sqrt{2}$$. Thus $$a_n = \sqrt{6}(\sqrt{2})^{n-1} = \sqrt{3}(\sqrt{2})^{n}$$.

##### Exercise 6
\begin{align*} \sum_{k=1}^n \Bigl(\dfrac{1}{k} - \dfrac{1}{k+1}\Bigr) &= \Bigl(1 - \dfrac{1}{2}\Bigr) + \Bigl(\dfrac{1}{2} - \dfrac{1}{3}\Bigr) + \Bigl(\dfrac{1}{3} - \dfrac{1}{4}\Bigr) + \dots + \Bigl(\dfrac{1}{n} - \dfrac{1}{n+1}\Bigr) \\ &= 1 - \dfrac{1}{n+1} = \dfrac{n}{n+1}. \end{align*}
##### Exercise 7

Writing the series forwards and backwards, we have

\begin{align} S_n = & a & + & (a+d) & + & (a+2d) & + & \cdots & + & (\ell-d) & + & \ell \\ S_n = & \ell & + & (\ell-d) & + & (\ell-2d) & + & \cdots & + & (a+d) & + & a. \end{align}

Adding in pairs gives

$2S_n = (a+\ell) + (a+\ell) + \dots + (a+\ell) = n(a+\ell).$

Hence, $$S_n = \dfrac{n}{2}(a+\ell)$$.

##### Exercise 8

Here $$a = \log_2 3$$ and $$d = \log_2 3$$, so

\begin{align*} S_n &= \dfrac{n}{2}\bigl(2\log_2 3 + (n-1)\log_2 3\bigr) = \dfrac{1}{2}n(n+1)\log_2 3. \end{align*}
##### Exercise 9

We have

\begin{align*} S_n &= a + ar + ar^2 + \cdots + ar^{n-1} \\ r S_n &= ar + ar^2 + \cdots + ar^{n-1} + ar^n. \end{align*}

Subtracting gives

\begin{align*} rS_n - S_n = ar^n - a \quad&\implies\quad S_n(r-1) = a(r^n-1) \\ &\implies\quad S_n = \dfrac{a(r^n-1)}{r-1}, \quad\text{provided } r\neq 1. \end{align*}
##### Exercise 10

We have $$a = \sqrt{3}$$ and $$r = \dfrac{6}{\sqrt{3}} = 2\sqrt{3}$$. Hence, $$S_n = \dfrac{\sqrt{3}\bigl((2\sqrt{3})^n - 1\bigr)}{2\sqrt{3} - 1}$$.

##### Exercise 11
1. The triangles $$AXY$$ and $$YXB$$ are similar. (They have equal angles, as $$\angle AYB = 90^\circ$$.) Thus $\dfrac{a}{XY} = \dfrac{XY}{b} \quad\implies\quad XY = \sqrt{ab}.$ Hence, the length $$XY$$ is the geometric mean of $$a$$ and $$b$$.
2. Let $$C$$ be the midpoint of $$AB$$. Let $$D$$ be the point where the perpendicular to $$AB$$ at $$C$$ cuts the semicircle. Then $$CD$$ is the radius of the semicircle, and so $$CD = \dfrac{1}{2}(a+b)$$. Clearly, $$CD \geq XY$$, and therefore $$\dfrac{1}{2}(a+b) \geq \sqrt{ab}$$.
##### Exercise 12

The common ratio is

$r = \dfrac{1}{1+\sqrt{2}} = \sqrt{2} - 1,$

and so $$-1 < r < 1$$. Hence, the geometric series has a limiting sum, given by

$S_\infty = \dfrac{1}{1 - (\sqrt{2} - 1)} = \dfrac{1}{2 - \sqrt{2}} = 1 + \dfrac{\sqrt{2}}{2}.$
##### Exercise 13

We can write the decimal $$0.\overline{12}$$ as

$0.\overline{12} = 0.12121212\ldots = \dfrac{12}{10^2} + \dfrac{12}{10^4} + \dfrac{12}{10^6} + \dotsb,$

which is a geometric series with $$a = \dfrac{12}{10^2}$$ and $$r = \dfrac{1}{10^2}$$. The limiting sum is

$0.\overline{12} = \Bigl( \dfrac{12}{10^2} \Bigr) \times \Bigl( \dfrac{1}{1 - \dfrac{1}{10^2}} \Bigr) = \dfrac{4}{33}.$
##### Exercise 14

The geometric mean of $$a_1,a_2,\dots,a_n$$ is $$G = \sqrt[n]{a_1a_2\dots a_n}$$. So

\begin{align*} \log_b G &= \log_b\bigl((a_1a_2\dots a_n)^{\frac{1}{n}}\bigr) \\ &= \dfrac{1}{n} \log_b(a_1a_2\dots a_n) \\ &= \dfrac{1}{n} \bigl(\log_b a_1 + \log_b a_2 + \dots + \log_b a_n\bigr), \end{align*}

which is the arithmetic mean of $$\log_b a_1,\ \log_b a_2,\ \dots,\ \log_b a_n$$.

##### Exercise 15
1. The arithmetic mean is $$AM = 4$$ and the geometric mean is $$GM = \sqrt{60} \approx 3.91$$. The harmonic mean $$HM$$ satisfies $\dfrac{1}{HM} = \dfrac{1}{3}\Bigl(\dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5}\Bigr) = \dfrac{47}{180} \quad\implies\quad HM = \dfrac{180}{47} \approx 3.83.$ So in this case $$HM \leq GM \leq AM$$.
2. Using the AM–GM inequality, we have $$a+b \geq 2\sqrt{ab}$$, and so $HM = \dfrac{2ab}{a+b} \leq \dfrac{2ab}{2\sqrt{ab}} = \sqrt{ab} = GM.$
##### Exercise 16
1. Let $$f(x) = \log_e x - x$$, for $$x > 0$$. Then $$f'(x) = \dfrac{1}{x} - 1$$, so the only stationary point is at $$x = 1$$. We could use a sign diagram to check that $$f(x)$$ has a maximum at $$x = 1$$. The maximum is $$f(1) = -1$$. So, for all $$x > 0$$, we have $$\log_e x - x \leq -1$$, giving $$\log_e x \leq x - 1$$.
2. Define $$A = \dfrac{1}{n}(a_1 + a_2 + \dots + a_n)$$. Substituting $$x = \dfrac{a_1}{A},\ x = \dfrac{a_2}{A},\ \dots,\ x = \dfrac{a_n}{A}$$ into the inequality $$\log_e x \leq x - 1$$ from part (a) gives \begin{align*} \log_e\Bigl(\dfrac{a_1}{A}\Bigr) &\leq \dfrac{a_1}{A} - 1 \\ \log_e\Bigl(\dfrac{a_2}{A}\Bigr) &\leq \dfrac{a_2}{A} - 1 \\ &\ \vdots\\ \log_e\Bigl(\dfrac{a_n}{A}\Bigr) &\leq \dfrac{a_n}{A} - 1. \end{align*} Adding, we have $\log_e\Bigl(\dfrac{a_1}{A}\Bigr) + \log_e\Bigl(\dfrac{a_2}{A}\Bigr) + \dots + \log_e\Bigl(\dfrac{a_n}{A}\Bigr) \leq \dfrac{1}{A}\bigl(a_1 + a_2 + \dots + a_n\bigr) - n.$ Hence, $\log_e\Bigl(\dfrac{a_1a_2\dots a_n}{A^n}\Bigr) \leq n-n = 0.$
3. Exponentiating both sides of the inequality above gives $\dfrac{a_1a_2\dots a_n}{A^n} \leq 1 \quad\implies\quad a_1a_2\dots a_n \leq A^n,$ from which we have $(a_1a_2\dots a_n)^{\dfrac{1}{n}} \leq A = \dfrac{a_1 + a_2 + \dots + a_n}{n},$ which is the desired result.