1 Solution 1

$(\dfrac{2n-1}{n})_{n=1}^\infty$

$a_n = \dfrac{2n-1}{n}$

$a_{n+1} = \dfrac{2(n+1)-1}{n+1} = \dfrac{2n+2-1}{n+1} = \dfrac{2n+1}{n+1}$

$a_{n+1} - a_n = \dfrac{2n+1}{n+1} - \dfrac{2n-1}{n}$

$= \dfrac{(2n+1)n - (2n-1)(n+1)}{(n+1)n}$

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

$= \dfrac{1}{(n+1)n}$

For $n \geq 1$, we know that $\dfrac{1}{n(n+1)} > 0$.

So, $a_{n+1} > a_n$

Hence we conclude that the sequence is strictly increasing.

2 Solution 2

$(\dfrac{n}{2n+1})_{n=1}^\infty$

$a_n = \dfrac{n}{2n+1}$

$a_{n+1} = \dfrac{n+1}{2(n+1)+1} = \dfrac{n+1}{2n+3}$

$a_{n+1}-a_n = \dfrac{n+1}{2n+3} - \dfrac{n}{2n+1}$

$= \dfrac{(n+1)(2n+1)-n(2n+3)}{(2n+3)(2n+1)}$

$= \dfrac{2n^2+n+2n+1-2n^2-3n}{(2n+3)(2n+1)}$

$= \dfrac{1}{(2n+3)(2n+1)}$

For $n \geq 1$, we know that $\dfrac{1}{(2n+3)(2n+1)} > 0$.

So, $a_{n+1} > a_n$

Hence we conclude that the sequence is strictly increasing.

3 Solution 3

$(n^2 - 2n)_{n=1}^\infty$

$a_n = n^2 - 2n$

$a_{n+1} = (n+1)^2 - 2(n+1)$

$= n^2 + 1 + 2n - 2n -2$

$= n^2 - 1$

$a_{n+1} - a_n = n^2 - 1 -(n^2 - 2n)$

$= 2n - 1$

For $n \geq 1$, we know that $2n-1 > 0$.

So, $a_{n+1} > a_n$.

Hence we conclude that the sequence is strictly increasing.

4 Solution 4

$(n^2-3n)_{n=1}^\infty$

$a_n = n^2 - 3n$

$a_{n+1} = (n+1)^2 - 3(n+1)$

$= n^2 + 2n + 1 -3n - 3$

$= n^2 - n - 2$

$a_{n+1}-a_n = n^2 - n - 2 - n^2 + 3n$

$= 2n - 2 = 2(n-1)$

For $n \geq 1, 2(n-1) \geq 0$

So, $a_{n+1} \geq a_n$

Hence we conclude that the sequence is weakly increasing.

5 Solution 5

$(n^2 - 4n)_{n=1}^\infty$

$a_n = n^2 - 4n$

$a_{n+1} = (n+1)^2 - 4(n+1)$

$= n^2 + 1 + 2n - 4n - 4$

$= n^2 + 1 - 2n -4 = n^2 - 2n -3$

$a_{n+1} - a_n = n^2 - 2n - 3 - n^2 + 4n$

$= 2n - 3$

When $n = 1, 2n - 3 < 0$.

When $n = 2, 2n - 3 > 0$.

So, the sequence diverges.

6 Solution 6

$(\dfrac{2^{3n+1}}{3^{2n+1}})_{n=1}^\infty$

$a_n = \dfrac{2^{3n+1}}{3^{2n+1}}$

$a_{n + 1} = \dfrac{2^{3(n+1)+1}}{3^{2(n+1)+1}} = \dfrac{2^{3n+4}}{3^{2n + 3}}$

$\dfrac{a_{n+1}}{a_n} = \dfrac{2^{3n+4}.3^{2n+1}}{2^{3n+1}.3^{2n+3}}$

$= 2^{3n + 4 - 3n -1}.3^{2n+1-2n-3}$

$= 2^3.3^{-2}$

$= \dfrac{8}{9}$

So, $\dfrac{a_{n+1}}{a_n} = \dfrac{8}{9}$

$\dfrac{a_n}{a_{n+1}} = \dfrac{9}{8}$

We know that $\dfrac{9}{8} > 1$

So, $a_n > a_{n+1}$

Hence we conclude that the sequence is strictly decreasing.

7 Solution 7

$(\dfrac{(2n)!}{4^n(n!)^2})_{n=1}^\infty$

$a_n = \dfrac{(2n)!}{4^n(n!)^2}$

$a_{n+1} = \dfrac{(2(n+1))!}{4^{n+1}((n+1)!)^2}$

$= \dfrac{(2n)!(2n+2)(2n+1)}{4.4^n((n+1)(n)!)^2}$

$\dfrac{a_{n+1}}{a_n} = \dfrac{(2n)!(2n+2)(2n+1)}{4.4^n((n+1)(n)!)^2} * \dfrac{4^n(n!)^2}{(2n)!}$

$= \dfrac{(2n+2)(2n+1)}{4(n+1)^2}$

$= \dfrac{2(2n+1)}{4(n+1)} = \dfrac{2n+1}{2(n+1)} = \dfrac{2n+1}{2n+2}$

So, $\dfrac{a_{n+1}}{a_n} = \dfrac{2n+1}{2n+2}$

Since $2n+2 > 2n+1$

$\dfrac{2n+2}{2n+1} > 1$

So, $a_{n+1} > a_n$

Hence we conclude that the sequence is strictly increasing.

8 Solution 8

$(\dfrac{(3n)!n!}{((2n)!)^2})_{n=1}^\infty$

$a_n = \dfrac{(3n)!n!}{((2n)!)^2}$

$a_{n+1} = \dfrac{(3(n+1)!(n+1)!}{((2(n+1))!)^2}$

$= \dfrac{(3n+3)(3n+2)(3n+1)(3n)!(n+1)n!}{((2n+2)(2n+1)(2n)!)^2}$

$\dfrac{a_{n+1}}{a_n} = \dfrac{(3n+3)(3n+2)(3n+1)(n+1)}{((2n+2)(2n+1))^2}$

$= \dfrac{3(n+1^2)(3n+1)(3n+2)}{(2n+2)^2(2n+1)^2}$

$= \dfrac{3(3n+1)(3n+2)}{4(2n+1)^2}$

We know that $2n+1 < 3n+1$ and also $2n + 1 < 3n + 2$

So, $(2n+1)^2 < (3n+1)(3n+2)$

So, $\dfrac{(3n+1)(3n+2)}{(2n+1)^2} > 1$

$\dfrac{3(3n+1)(3n+2)}{4(2n+1)^2} > \dfrac{3}{4}$

So, $\dfrac{a_{n+1}}{a_n} > \dfrac{3}{4}$

$\dfrac{a_n}{a_{n+1}} > \dfrac{4}{3}$

So, $a_n > a_{n+1}$

Hence we conclude that the sequence is strictly decreasing.

9 Solution 9

$(\sin(\dfrac{n\pi}{4}))_{n=1}^\infty$

$a_n = \sin(\dfrac{n\pi}{4})$

$a_{n+1} = \sin(\dfrac{(n+1)\pi}{4}) = \sin (\dfrac{n\pi}{4} + \dfrac{\pi}{4})$

$a_1 = \sin (\dfrac{\pi}{4})$

$a_2 = \sin (\dfrac{2\pi}{4})$

$a_3 = \sin (\dfrac{3\pi}{4})$

$a_4 = \sin (\dfrac{4\pi}{4})$

Since the $\sin$ function oscillates up and down between $-1$ and $1$ as $n \to \infty$, we can say that the sequence diverges.

10 Solution 10

$(\sin (\dfrac{\pi}{4n}))_{n=1}^\infty$

$a_n = \sin (\dfrac{\pi}{4n})$

$\lim_{x \to \infty} \dfrac{\pi}{4x} = 0$

$\lim_{x \to \infty} \sin (\dfrac{\pi}{4x}) = 0$

From theorem 2.8.2, $\lim_{x \to \infty} \sin (\dfrac{\pi}{4x}) = 0$

From theorem 2.9.5, we can deduce that the sequence is monotone.

11 Solution 11

$(a_n)_{n=1}^\infty$

$(a_n) = (\dfrac{9}{10})^n$

11.1 Solution a

To prove $a_{n+1} = \dfrac{9}{10}a_n$

We will use mathematical induction.

Base case: $n=1$

$a_{1+1} = a_2 = (9/10)^2 = (9/10).a_1$

Induction step: Suppose $n$ is a positive integer and $a_{n+1} = \dfrac{9}{10}a_n$

$a_{n+2} = \dfrac{9}{10}^{n+2}$

$= (\dfrac{9}{10})^n (\dfrac{9}{10})^2$

$= (\dfrac{9}{10}(\dfrac{9}{10})^{n+1})$

$= (\dfrac{9}{10}).a_{n+1}$

This concludes the proof.

11.2 Solution b

$a_{n+1} = \dfrac{9}{10}.a^n$

$\dfrac{a_{n+1}}{a_n} = \dfrac{9}{10}$

$\dfrac{a_n}{a_{n+1}} = \dfrac{10}{9}$

Since $\dfrac{10}{9} > 1, \dfrac{a_n}{a_{n+1}} > 1$. So, $a_n > a_{n+1}$.

Hence we conclude that the sequence is strictly decreasing.

11.3 Solution c

Since the sequence is strictly decreasing, then from theorem 2.9.5, eitehr the sequence converges or diverges.

The first few terms of the sequence are

$a_1 = 9/10, a_2 = (9/10^2) = 0.81, a_3 = (9/10)^3 = 0.729$

Let's try to prove for every integer $n$,

$1 > a_n > a_{n+1} > 0$

Proving this we can conclude that $0$ is a lower bound for the sequence.

We will use mathematical induction to prove it.

Base case. Since $a_1 = 9/10$ and $a_2 = (9/10)^2$, we have $1 > 9/10 > (9/10)^2 > 0$

Induction step: Suppose $n$ is a positive integer and $1 > a_n > a_{n+1} > 0$.

$\dfrac{9}{10} > \dfrac{9}{10}a_n > \dfrac{9}{10}a_{n+1} > 0$

$1 > \dfrac{9}{10} > a_{n+1} > a_{n+2} > 0$

$1 > a_{n+1} > a_{n+2} > 0$

This completes the mathematical induction proof.

11.4 Solution d

Since the sequence converges, there is some number $L$ such that as $n \to \infty, a_n \to L$.

The number $L$ is the least upper bound for the set of terms of the sequence.

We know that $\lim_{n \to \infty} a_{n+1} = L$.

So, $\lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \dfrac{9}{10}a_n$

$L = \dfrac{9}{10}L$

$10L-9L = 0$

$L = 0$

12 Solution 12

$(a_n)_{n=1}^\infty$

$a_n = (11/10)^n$

12.1 Solution a

$a_{n+1} = (\dfrac{11}{10}).a_n$

We will use mathematical induction to prove it.

Base case: $n=1$

$a_2 = (11/10)^2 = (11/10).a_1$

Induction step: Suppose $n$ is a positive integer and $a_{n+1} = (\dfrac{11}{10}).a_n$

$a_{n+2} = (11/10)^{n+2} = (11/10).(11/10)^{n+1}$

$= (11/10).a_{n+1}$

This completes the mathematical induction proof.

12.2 Solution b

$a_{n+1} = (11/10)a_n$

$\dfrac{a_{n+1}}{a_n} = \dfrac{11}{10}$

Since $11/10 > 1, a_{n+1}/a_n > 1$.

So, $a_{n+1} > a_n$

Hence we conclude that the sequence is strictly increasing.

12.3 Solution c

From theorem 2.9.5, we know that the sequence either converges or $\lim_{n \to \infty} a_n = \infty$

Suppose that the sequence converges - So, $\lim_{n \to \infty} a_n = L$. Then $L$ is the least upper bound for the set of terms of the sequence. Since we know that $a_4 = (11/10)^4$, we must have $L \geq (11/10)^4$.

Now we can compute $\lim_{n \to \infty} a_{n+1}$ in two ways. As $n \to \infty$, we have $n+1 \to \infty$ and therefore $\lim_{n \to \infty} = L$. Also,

$a_{n+1} = (11/10)a_n$

$\lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} (11/10)a_n = \dfrac{11}{100}L$

So, $L = 11/10 . L$

Dividiving by L, we get $1 = 11/10$ which is obviously false. We conclude that our assumption that the sequence convertes must be wrong.

Thus, the sequence does not converge. By theorem 2.9.5, the only remaining possibility is that $\lim_{n \to \infty} a_n = \infty$

13 Solution 13

$(a_n)_{n=1}^\infty$

$a_n = \dfrac{3^n}{(n+2)!}$

13.1 Solution a

We will prove using mathematical induction.

Base case: $n = 1$

$a_2 = \dfrac{3^2}{4!} = \dfrac{3.3}{4.3!} = \dfrac{3.3}{(1+3)3!}$

$= \dfrac{3.3!}{(1+3)(1+2)!} = \dfrac{3.a_1}{(1+3)}$

Induction step: Suppose $n$ is a positive integer and $a_{n+1} = (\dfrac{3}{n+3}).a_n$

$a_{n+2} = \dfrac{3^{n+2}}{(n+4)!} = \dfrac{3^n.3^2}{(n+4)(n+4)(n+2)!}$

$= \dfrac{a^n.3^2}{(n+4)(n+3)}$

$= \dfrac{a_{n+1}.3}{n+4}$

$= \dfrac{3.a_{n+1}}{((n+1)+3)}$

This completes the mathematical induction proof.

13.2 Solution b

$\dfrac{a_{n+1}}{a_n} = \dfrac{3}{n+3}$

Since $n$ is a positive integer $\geq 1, n + 3 > 3$.

So, $\dfrac{n+3}{3} > 1$. So, $\dfrac{a_{n+1}}{a_n} < 1$.

$a_{n+1} < a_n$

$a_n > a_{n+1}$

So the sequence is strictly decreasing.

13.3 Solution c

The first few terms of the sequence are

$a_1 = \dfrac{3}{3!} = \dfrac{1}{2}, a_2 = \dfrac{3^2}{4!} = \dfrac{3}{8}$

$a_3 = \dfrac{3^3}{5!} = \dfrac{9}{40}$

Let's try to prove for every integer $n$, $a > a_n > a_{n+1} > 0$.

Proving that we can infer that $0$ is the lower bound for the sequence.

We will use mathematical induction to prove it.

Base case: $n=1$

$a_1 = \dfrac{1}{2}$ and $a_2 = \dfrac{3}{8}$

So, $1 > \dfrac{1}{2} > \dfrac{3}{8} > 0$

Induction step: Suppose $n$ is a positive integer and $1 > a_n > a_{n+1} > 0$.

$\dfrac{3}{n+3} > \dfrac{3}{n+3}.a_n > \dfrac{3}{n+3}.a_{n+1} > 0$

We know $n + 4 > n + 3$

$\dfrac{1}{n+3} > \dfrac{1}{n+4}$

$\dfrac{3}{n+3}a_{n+1} > \dfrac{3}{n+4}a_{n+1}$

So, $\dfrac{3}{n+3} > \dfrac{3}{n+3}a_n > \dfrac{3}{n+4}a_{n+1} > 0$

We know that $a_{n+1} = \dfrac{3}{n+3}a_n$ and $a_{n+2} = \dfrac{3}{n+4}a_{n+1}$

So, $\dfrac{3}{n+3} > a_{n+1} > a_{n+2} > 0$

We know that $n+3 > 3$

$1 > \dfrac{3}{n+3}$

So, $1 > \dfrac{3}{n+3} > a_{n+1} > a_{n+2} > 0$

$1 > a_{n+1} > a_{n+2} > 0$

This completes the mathematical induction proof.

13.4 Solution d

Since the sequence converges, there is some number $L$ such that as $n \to \infty, a_n \to L$.

The number $L$ is the greatest lower bound for the set of terms of the sequence. Since $a_3 = \dfrac{9}{40}$ and $0$ is the lower bound, we must have $0 \leq L \leq \dfrac{9}{40}$.

We are going to compute $\lim_{n \to \infty} a_{n+1}$ in two different ways and then set the two answers equal to each other. For the first computation and let $m = n+1$. Then as $n \to \infty, m=n+1 \to \infty$ and as $m \to \infty, a_m \to L$. Combining these, we conclude that as $n \to \infty, a_{n+1}=a_m \to L$. So, $\lim_{n \to \infty}a_{n+1} = L$. For the second computation, we use the recursive definition of the sequence to see that

$\lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \dfrac{3}{n+3}a_n = 0$

Both computation are correction, so the answers must be equal to each other. Therefore $L = 0$.

14 Solution 14

14.1 Solution a

$(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are strictly decreasing.

So, $a_n > a_{n+1}$ and $b_n > b_{n+1}$

We need to prove that $a_n + b_n > a_{n+1} + b_{n+1}$

So, $a_n - a_{n+1} > 0$

$b_n - b_{n+1} > 0$

$a_n + b_n - a_{n+1} - b_{n+1} > 0$

$a_n + b_n > a_{n+1} + b_{n+1}$

This concludes that the series $(a_n + b_n)_{n=1}^\infty$ is strictly decreasing.

14.2 Solution b

$a_{n+1} - a_n = \dfrac{3^{n+1}+4^{n+1}}{7^{n+1}} - \dfrac{3^n + 4^n}{7^n}$

$= \dfrac{3^{n+1}+4^{n+1}-3^n.7-4^n.7}{7^{n+1}}$

$= \dfrac{3^n.3 + 4^n.4 - 3^n.7 - 4^n.7}{7^{n+1}}$

Now we know that

$7 > 3$

$3^n.7 > 3^n.3$

$0 > 3^n.3 - 3^n.7$

Similarly, $0 > 4^n.4 - 4^n.7$

Combining them,

$0 > 3^n.3 - 3^n.7 + 4^n.4 - 4^n.7$

$0 > \dfrac{3^n.3 - 3^n.7 + 4^n.4 - 4^n.7}{7^{n+1}}$

So, $0 > a_{n+1} - a_n$

$a_n > a_{n+1}$

So the sequence is strictly decreasing.

15 Solution 15

$a_1 = 2, a_{n+1} = 3 + \dfrac{a_n}{2}$

15.1 Solution a

$a_{n+1} > a_n$

We will prove using mathematical induction.

Base case: $n=1$

$a_2 = 3 + \dfrac{2}{2} = 4$

$a_1 = 2$

$a_2 > a_1$

Induction step: Suppose $n$ is a positive integer and $a_{n+1} > a_n$.

$a_{n+2} = 3 + \dfrac{a_{n+1}}{2}$

$a_{n+1} > a_n$

$\dfrac{a_{n+1}}{2} > \dfrac{a_n}{2}$

$3 + \dfrac{a_{n+1}}{2} > 3 + \dfrac{a_n}{2}$

$a_{n+2} > 3 + \dfrac{a_n}{2}$

$a_{n+2} > a_{n+1}$

This completes the mathematical induction proof.

15.2 Solution b

$a_n < 10$

We will prove using mathematical induction.

Base case: $n=1$

$a_1 = 2 < 10$

Induction case: Suppose $n$ is a positive integer and $a_n < 10$

$a_n < 10$

$\dfrac{a_n}{2} < 5$

$3 + \dfrac{a_n}{2} < 8$

$a_{n+1} < 8 < 10$

So, $a_{n+1} < 10$

This completes the mathematical induction proof.

15.3 Solution c

$\lim_{n \to \infty} a_n$

From (a) we know that the sequence is strictly increasing.

Let's assume that the sequence converges. So, $\lim_{n \to \infty} a_n = L$. Also, we know that $\lim_{n \to \infty} a_{n+1} = L$.

$\lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} 3 + \dfrac{a_n}{2} = 3 + \dfrac{L}{2}$

$L = 3 + \dfrac{2}{2}$

$L = 6$

16 Solution 16

$(a_n)_{n=1}^\infty$

$a_1 = 2, a_{n+1} = \sqrt{8 + \dfrac{a_n^2}{2}}$

16.1 Solution a

$a_{n+1} > a_n$

We will prove using mathematical induction.

Base case: $n=1$

$a_1 = 2$

$a_2 = \sqrt{8 + \dfrac{2^2}{2}} = \sqrt{8+2} = \sqrt{10}$

$a_2 > a_1$

Induction case: Suppose $n$ is a positive integer and $a_{n+1} > a_n$

So, $a_{n+1} > a_n$

$a_{n+1}^2 > a_n^2$

$\dfrac{a_{n+1}^2}{2} > \dfrac{a_n^2}{2}$

$8 + \dfrac{a_{n+1}^2}{2} > 8 + \dfrac{a_n^2}{2}$

$\sqrt{8 + \dfrac{a_{n+1}^2}{2}} > \sqrt{8 + \dfrac{a_n^2}{2}}$

$a_{n+2} > a_{n+1}$

This completes the mathematical induction proof.

16.2 Solution b

We will prove that $0 < a_n < a_{n+1} < 4$. This will prove that $4$ is the upper bound of the sequence.

Base case: $n=1$

$a_1 = 2$

$a_2 = \sqrt{10}$

$0 < a_1 < a_2 < 4$

Induction case: Suppose $n$ is a positive integer and $0 < a_n < a_{n+1} < 4$.

From induction step,

$0 < a_n < a_{n+1} < 4$

$0 < a_n^2 < a_{n+1}^2 < 16$

$0 < \dfrac{a_n^2}{2} < \dfrac{a_{n+1}^2}{2} < 8$

$8 < 8 + \dfrac{a_n^2}{2} < 8 + \dfrac{a_{n+1}^2}{2} < 16$

$\sqrt{8} < \sqrt{8 + \dfrac{a_n^2}{2}} < \sqrt{8 + \dfrac{a_{n+1}^2}{2}} < 4$

$\sqrt{8} < a_{n+1} < a_{n+2} < 4$

Since $0 < \sqrt{8}$

$0 < a_{n+1} < a_{n+2} < 4$

This completes the mathematical induction proof.

16.3 Solution c

From (a), we know that the sequence is strictly increasing. From (b) we know that the sequence has an upper bound. So, from theorem 2.9.5, we know that the sequence onverges.

So, $\lim_{n \to \infty} a_n = L$

Also, we know that $\lim_{n \to \infty} a_{n+1} = L$

$\lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \sqrt{8 + \dfrac{a_n^2}{2}}$

$= \lim{n \to \infty} \sqrt{\dfrac{16 + a_n^2}{2}}$

$L = \sqrt{\dfrac{16 + L^2}{2}}$

$2L^2 = 16$

$L = 4$

17 Solution 17

$(a_n)_{n=1}^\infty$

$a_1 = 1, a_{n+1} = 6 - \dfrac{9}{a_n}$

17.1 Solution a

$a_1 = 1$

$a_2 = 6 - \dfrac{9}{1} = -3$

$a_3 = 6 - \dfrac{9}{-3} = 6 + 3 = 9$

$a_4 = 6 - \dfrac{9}{9} = 6 - 1 = 5$

17.2 Solution b

$\forall n \geq 3 \; a_n > a_{n+1} > 3$

We will prove using mathematical induction.

Base case: $n=3$

$a_3 = 9$

$a_4 = 5$

$a_3 > a_4 > 3$

Induction case: Suppose $n$ is a positive integer $\geq 3$ and $a_n > a_{n+1} > 3$.

$a_n > a_{n+1}$

$\dfrac{1}{a_{n+1}} > \dfrac{1}{a_n}$

$\dfrac{-9}{a_{n+1}} < \dfrac{-9}{a_n}$

$6 - \dfrac{-9}{a_{n+1}} < 6 - \dfrac{-9}{a_n}$

$a_{n+2} < a_{n+1}$

We know that $a_{n+1} > 3$

$1 > \dfrac{3}{a_{n+1}}$

$3 > \dfrac{9}{a_{n+1}}$

$-3 < \dfrac{-9}{a_{n+1}}$

$3 < 6 \dfrac{-9}{a_{n+1}}$

$3 < a_{n+2}$

Combining with above,

$3 < a_{n+2} < a_{n+1}$

$a_{n+1} > a_{n+2} > 3$

This completes the mathematical induction proof.

17.3 Solution c

From (b) we know that the sequence is strictly decreasing and that it has a lower bound. From theorem 2.9.5, we know that the sequence converges.

So, $\lim_{n \to \infty} a_n = L$

Also, we know that $\lim_{n \to \infty} a_{n+1} = L$

$\lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} 6 - \dfrac{9}{a_n} = 6 - \dfrac{9}{L}$

$L = 6 - \dfrac{9}{L}$

$L^2 = 6L - 9$

$L^2 - 6L + 9 = 0$

$L^2 - 3L - 3L + 9 + 0$

$(L-3)^2 = 0$

$L = 3$

18 Solution 18

We know from example 2.9.7 that the sequence is strictly increasing.

By proving $\forall n \; \dfrac{(2n)!}{(n!)^2} \geq 2^n$, we will conclude that the sequence does not have upper bound. Then using the theorem 2.9.5, we can conclude that the limit of the sequence is infinity.

We will use mathematical induction to prove it.

Base case: $n=1$

$\dfrac{(2n)!}{(n!)^2} = \dfrac{2!}{1^2} = 2$

$2^n = 2$

So, $2 \geq 2$

Induction case: Suppose $n$ is a positive integer and $\dfrac{(2n)!}{(n!)^2} \geq 2^n$

$\dfrac{(2(n+1))!}{((n+1)!)^2} = \dfrac{(2n+2)(2n+1)(2n)!}{(n+1)^2(n!)^2}$

$= \dfrac{2(2n+1)(2n)!}{(n+1)(n!)^2}$

We know $\dfrac{(2n)!}{(n!)^2} \geq 2^n$

$\dfrac{2(2n+1)(2n)!}{(n+1)(n!)^2} \geq \dfrac{2(2n+1)2^n}{(n+1)}$

$\dfrac{(2(n+1))!}{((n+1)!)^2} \geq 2^{n+1}\dfrac{2n+1}{n+1}$

We know that

$2n+1 > n+1$

$\dfrac{2n+1}{n+1} > 1$

$\dfrac{(2n+1)(2^{n+1})}{n+1} > 2^{n+1}$

So, $\dfrac{(2(n+1))!}{((n+1)!)^2} \geq \dfrac{2^{n+1}(2n+1)}{n+1} > 2^{n+1}$

$\dfrac{(2(n+1))!}{((n+1)!)^2} > 2^{n+1}$

This completes the mathematical induction proof.

19 Solution 19

$\dfrac{n!}{k^n}$

$a_n = \dfrac{n!}{k^n}$

Let's find the property of $(a_n)_{n=k}^\infty$

$a_k = \dfrac{k!}{k^k}$

$a_{k+1} = \dfrac{(k+1)!}{k^{k+1}}$

$\dfrac{a_{k+1}}{a_k} = \dfrac{(k+1)!k^k}{k^{k+1}.k!} = \dfrac{k+1}{k}$

$k + 1 > k$

$\dfrac{k+1}{k} > 1$

So, $\dfrac{a_{k+1}}{a_k} > 1$

$a_{k+1} > a_k$

So the sequence $(a_n)_{n=k}^\infty$ is strictly increasing.

Thus according to theorem 2.9.5, either the sequence converges or diverges. Let's assume that the sequence converges. Then $L$ is the upper bound for the set of the terms of the sequence.

We know that $\lim_{n \to \infty} a_{n+1} = L$

$a_{n+1} = \dfrac{(n+1)!}{k^{n+1}} = \dfrac{(n+1)a^n}{k}$

$\lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \dfrac{(n+1)a^n}{k} = \infty$

$L = \infty$

This contradics our assumption that a limit exists. Hence the sequence diverges.

20 Solution 20

20.1 Solution a

Suppose $(a_n)_{n=1}^\infty$ is weakly decreasing and $A$ has a lower bound. Then by completenss of real numbers, $A$ has a greatest lower bound. Let $b$ be the greatest lower bound. We now use the definition of limits to show that $\lim_{n \to \infty} a_n = b$.

Suppose $\epsilon > 0$. Since $b$ is the greatest lower bound of $A$, $b + \epsilon$ is not an lower bound, so there must be some positive integer $N$ such that $a_N < b + \epsilon$.

Now consider any integer $n > N$. Since the sequence is weakly decreasing $a_n \leq a_N$, So, $a_n \leq a_N < b + \epsilon$. And since $L$ is an lower bound for $A$, $a_n \geq b$. Let's combine these facts,

$b \leq a_n < b + \epsilon$

$b - \epsilon < b \leq a_n < b + \epsilon$

$b - \epsilon < a_n < b + \epsilon$

Thus for every $n > N$ we have $|a_n - b|$ as required.

20.2 Solution b

Suppose $A$ does not have lower bound. Let us consider a number $M$.

Since $M$ is not a lower bound there must be some positive integer $N$ such that $a_N < M$.

Since the sequence is weakly decreasing for any integer $n < N, a_n \leq a_N < M$, so for every number $M$, there is some number $N$ such taht $n > N$ then $a_n < M$.

So, $\lim_{n \to \infty} a_n = -\infty$

21 Solution 21

Suppose that $A$ has an lower bound. Let $u_1$ be an lower bound for $A$. The assumption that $A$ is non empty means that there is at least one number in $A$, so we can let $a$ be some element of $A$. Then if we let $v_1$ be any number greater than $a$, then $v_1$ is not an lower bound for $A$.Note that $u_1 \leq a < v_1$. We ill look for the greatest lower bound between $u_1$ and $v_1$. We will be using the following lemma to narrow down the search.

Lemma: Suppose $u < v$. $u$ is an lower bound for $A$ and $v$ is not an lower bound for $A$. Then there are numbers $u'$ and $v'$ such that $u \leq u' < v' \leq v, v'-u'=\dfrac{v-u}{2}$, $u'$ is an lower bound for $A$ and $v'$ is not an lower bound for $A$.

Proof of lemma: Let $m = (u+v)/2$, the midpoint of the interval $[u,v]$. If $m$ is the lower bound for $A$, then we can let $u'=m, v'=v$. If not, $u'=u$ and $v'=m$.

By the above lemma, there must be number $u_2$ and $v_2$ such that $u_1 \leq u_2 < v_2 \leq v_1, v_2 - v_1 = \dfrac{v_1 - u_1}{2}$, $v_2$ is not the lower bound and $u_2$ is an lower bound.

Applying the lemma again, we see that there are numbers $u_3$ and $v_3$ such that $u_2 \leq u_3 < v_3 \leq v_2, v_3 - u_3 = \dfrac{(v_2 - u_2)}{2} = \dfrac{(v_1 - u_1)}{4}$, $u_3$ is an lower bound for $A$ and $v_3$ is not an lower bound.

Continuing in this way, we get sequence $(u_n)_{n=1}^\infty$ and $(v_n)_{n=1}^\infty$ such that $u_1 \leq u_2 \leq \leq u_3 < ...\cdot$, $v₁ ≥ v₂ ≥ v₃ ≥ …$ and for every $n$, $u_n \leq v_n, v_n - u_n = \dfrac{v_1 - u_1}2^{n-1}$, $u_n$ is an lower bound of $A$ and $v_n$ is not an lower bound for $A$. Since $\lim_{n \to \infty} v_n - u_n = \lim_{n \to \infty} \dfrac{v_1 - u_1}{2^n - 1} = 0$, the nested interval theorem implies that there is a number $b$ such that for every $n, u_n \leq b \leq v_n$ and $\lim_{n \to \infty} u_n = \lim_{n \to \infty} v_n = b$. We will show that $b$ is the greatest lower bound for $A$.

To see that $b$ is an lower bound for $A$. Let $x$ be any element of $A$. For every $n$, $u_n$ is a lower bound for $A$, so $x \geq u_n$. Since weak inequalities are presereved by limits, it follows that $b = \lim_{n \to \infty} u_n \leq x$. So, $b$ is an lower bound. To see that $b$ is the greatest lower bound, consider any number $c>b$. Since $\lim_{n \to \infty} v_n = b$, by the sequence version of theorem 2.8.8, for sufficiently large $n$, $c > v_n$. Since $v_n$ is not an lower bound for $A$, there is some $x \in A$ such that $x < v_n$. Combininig with $v_n < c$, we have $x < c$, so $c$ is not an lower bound for $A$. Thus, we see that no number greater than $b$ can be a lower bound for $A$. So, $b$ is the greatest lower bound of $A$.