Sequences

Lessons

Senior 5 · Mathematics

Sequences

4 lessons

Lesson 4 of 4

Convergence and sequences in real life

25 min

Where is the sequence going?

Start with $u_{1} = 1$ and repeat the rule $u_{n+1} = \frac{1}{2}\left(u_n + \frac{2}{u_n}\right)$ , a method used in ancient Babylon:

$n$	1	2	3	4	5
$u_{n}$	1	1.5	1.416 667	1.414 216	1.414 214

After four steps the terms agree with $\sqrt{2} = 1.414\,213\,6\ldots$ to six decimal places, and later terms stay there. We say the sequence converges to $\sqrt{2}$ . Other sequences, such as $2, 4, 8, 16, \dots$ or $1, -1, 1, -1, \dots$ , never settle. This lesson makes "settles down" precise, gives rules for finding limits, and then uses sequences to answer real questions.

Definition

Convergent sequence: $(u_{n})$ converges to the real number $L$ , written $\lim_{n \to \infty} u_n = L$ or $u_n \to L$ , if the terms become and stay as close to $L$ as we wish: for every $\varepsilon \gt 0$ there is a number $N$ such that

|u_n - L| \lt \varepsilon \quad \text{for every } n \gt N.

$L$ is the limit. A sequence that does not converge is divergent; it may tend to $+\infty$ or $-\infty$ , or oscillate.

Example: $\frac{1}{n} \to 0$ : for $\varepsilon = 0.001$ , every term after $N = 1000$ is within 0.001 of 0.

Worked example 1: finding $N$

Show that $u_n = \frac{2n - 1}{n + 3}$ (from Lesson 1) converges to 2, and find $N$ such that $|u_n - 2| \lt 0.01$ for every $n \gt N$ .

Solution $u_n - 2 = \frac{2n - 1 - 2n - 6}{n + 3} = -\frac{7}{n + 3}$ , so $|u_n - 2| = \frac{7}{n + 3}$ . This is less than any $\varepsilon \gt 0$ once $n + 3 \gt \frac{7}{\varepsilon}$ , so $u_n \to 2$ . For $\varepsilon = 0.01$ : $n + 3 \gt 700$ , so $N = 697$ works.

Rules for limits

If $u_n \to a$ and $v_n \to b$ , then $u_n + v_n \to a + b$ , $u_n v_n \to ab$ , $k\,u_n \to ka$ and, when $b \ne 0$ , $\frac{u_n}{v_n} \to \frac{a}{b}$ . With these and a few standard limits, most sequences can be handled:

Sequence	Behaviour as $n \to \infty$
$\frac{1}{n^p}$ , $p \gt 0$	converges to 0
$r^{n}$ , $\|r\| \lt 1$	converges to 0
$r^{n}$ , $r = 1$	constant 1, converges to 1
$r^{n}$ , $r \gt 1$	diverges to $+\infty$
$r^{n}$ , $r \le -1$	diverges, oscillating
arithmetic, $d \ne 0$	diverges to $\pm\infty$

Rational sequences: divide the top and the bottom by the highest power of $n$ in the denominator. Squeeze rule: if $a_n \le u_n \le b_n$ and $a_n, b_n \to L$ , then $u_n \to L$ . Monotone sequences: an increasing sequence that is bounded above converges, and so does a decreasing sequence that is bounded below (this is a property of the real numbers, proved at university).

Figure 1. The common ratio decides everything:

|r| \lt 1

gives a sequence that tends to 0 (alternating in sign if

r \lt 0

);

r \gt 1

gives one that grows without bound.

Worked example 2: four limits

Find (a) $\lim \frac{3n^2 - 5n}{2n^2 + 1}$ (b) $\lim \frac{2^n + 5^n}{5^n}$ (c) $\lim \frac{(-1)^n}{n}$ (d) $\lim \left(\sqrt{n + 1} - \sqrt{n}\right)$ .

Solution

Divide by $n^{2}$ : $\frac{3 - \frac{5}{n}}{2 + \frac{1}{n^2}} \to \frac{3 - 0}{2 + 0} = \frac{3}{2}$ .
$\frac{2^n + 5^n}{5^n} = \left(\frac{2}{5}\right)^n + 1 \to 0 + 1 = 1$ , since $\left|\frac{2}{5}\right| \lt 1$ .
$-\frac{1}{n} \le \frac{(-1)^n}{n} \le \frac{1}{n}$ , and both bounds tend to 0, so the limit is 0 by the squeeze rule.
Multiply by the conjugate: $\sqrt{n + 1} - \sqrt{n} = \frac{(n + 1) - n}{\sqrt{n + 1} + \sqrt{n}} = \frac{1}{\sqrt{n + 1} + \sqrt{n}} \to 0$ .

Warning

" $\infty - \infty$ " is not 0. In (d), both $\sqrt{n + 1}$ and $\sqrt{n}$ grow without bound, so you cannot subtract limits: rewrite first. In the same way $\frac{\infty}{\infty}$ has no fixed value; $\frac{n^2}{n} \to \infty$ but $\frac{n}{n^2} \to 0$ . A bounded sequence need not converge either: $(- 1)^{n}$ is bounded and divergent.

Sequences in real life

To model a situation with a sequence, decide what changes each step: a fixed amount added (arithmetic), a fixed percentage (geometric), or both (a recurrence $u_{n+1} = ru_n + c$ ). Then ask the question the situation needs: one term, a sum, or the long-run limit.

Worked example 3: a fish pond

A fish farmer's pond holds 2,000 tilapia. Each year 30% of the fish are harvested and then 500 young fish are added (illustrative figures). So $u_{n+1} = 0.7u_n + 500$ , $u_{0} = 2000$ . What happens in the long run?

Solution If the stock settles at $L$ , then $L = 0.7 L + 500$ , so $L = \frac{500}{0.3} \approx 1667$ . To prove that it does settle, subtract the two equations: $u_{n+1} - L = 0.7(u_n - L)$ . So $v_{n} = u_{n} - L$ is geometric with ratio 0.7, and

u_n = L + 0.7^n\,(u_0 - L) = 1666.7 + 333.3 \times 0.7^n \to 1666.7.

The pond settles at about 1,667 fish, from any starting stock (Figure 2).

Figure 2.

u_{n+1} = 0.7u_n + 500

(illustrative figures). Whatever the starting stock, the sequence converges to the same limit

L = \frac{500}{0.3} \approx 1667

. (Vertical axis starts at 300.)

Worked example 4: a pendulum dies away

A pendulum bob travels 50 cm on its first swing, and each swing is 90% as long as the one before. (a) How far has the bob travelled after 10 swings? (b) What total distance can it never exceed? (c) Which is the first swing shorter than 5 cm?

Solution The swings form a geometric sequence with $u_{1} = 50$ and $r = 0.9$ .

$D_{10} = 50 \times \frac{1 - 0.9^{10}}{1 - 0.9} = 500\,(1 - 0.3487) \approx 325.7$ cm.
$D_{n} = 500 (1 - {0.9}^{n})$ , and $0.9^n \to 0$ , so $D_n \to 500$ : the bob travels less than 5 m in total, however long it swings.
$50 \times 0.9^{n-1} \lt 5$ means $0.9^{n-1} \lt 0.1$ . Since $0.9^{21} \approx 0.109$ and $0.9^{22} \approx 0.098$ , we need $n - 1 = 22$ : the 23rd swing.

Worked example 5: a stack of bricks

Bricks at a building site are stacked in 15 rows: 40 in the bottom row and 2 fewer in each row above (illustrative figures). How many bricks are in the top row and in the whole stack?

Solution Arithmetic with $u_{1} = 40$ , $d = - 2$ : the top row has $u_{15} = 40 - 28 = 12$ bricks, and the stack has $S_{15} = \frac{15}{2}(40 + 12) = 390$ bricks.

Tip

The recurrence $u_{n+1} = ru_n + c$ (with $r \ne 1$ ) always has the fixed point $L = \frac{c}{1 - r}$ , and $u_{n} - L$ is geometric with ratio $r$ . So it converges to $L$ exactly when $|r| \lt 1$ . Savings with regular deposits, loans with regular repayments and medicine levels in the body all follow this rule.

The Organic Chemistry Tutor: Converging and Diverging Sequences Using Limits (practice problems). Watch how each limit is found by dividing by the highest power of n, and how a limit that does not exist means the sequence diverges.

✦

Practice

Figures in questions 5 to 7 are illustrative. The last two are harder.

Find the limits: (a) $\frac{3n^2 - 1}{n^2 + 4n}$ (b) $\frac{5n + 2}{n^2 + 1}$ (c) $\frac{2(-1)^n}{n}$ (d) $\frac{2^n + 1}{2^n}$ .
Converge or diverge? (a) $7 - 3 n$ (b) $(- 2)^{n}$ (c) ${0.9}^{n}$ (d) $\cos(n\pi)$ .
For $u_n = \frac{n + 1}{n}$ , find $N$ such that $|u_n - 1| \lt 0.001$ for all $n \gt N$ .
$u_{1} = 2$ and $u_{n+1} = \frac{1}{2}u_n + 3$ . Find $u_{2}$ , $u_{3}$ , $u_{4}$ , the limit $L$ , and a formula for $u_{n}$ .
A tree nursery plants 120 seedlings in its first month and 15 more each month than in the month before. How many does it plant in its first year?
A coffee cooperative produces 120 tonnes in year 1, and production rises by 8% a year. Find the production in year 6 and the total over the six years.
A reservoir loses 10% of its water each week, and 2,000 m³ is pumped in at the end of each week. Write the recurrence and find the long-run volume.
Apply $u_{n+1} = \frac{1}{2}\left(u_n + \frac{5}{u_n}\right)$ with $u_{1} = 2$ to find $u_{2}$ and $u_{3}$ . Show that if the sequence converges to $L \gt 0$ , then $L = \sqrt{5}$ .

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(a) 3 (b) 0 (c) 0, by the squeeze rule (d) $1 + \left(\frac{1}{2}\right)^n \to 1$ .
(a) diverges to $-\infty$ (b) diverges, oscillating with growing size (c) converges to 0 (d) $\cos n\pi = (-1)^n$ : diverges, oscillating.
$|u_n - 1| = \frac{1}{n} \lt 0.001$ when $n \gt 1000$ : $N = 1000$ .
$4, 5, 5.5$ ; $L = \frac{1}{2}L + 3$ gives $L = 6$ ; $u_n - 6 = (2 - 6)\left(\frac{1}{2}\right)^{n-1}$ , so $u_n = 6 - 4\left(\frac{1}{2}\right)^{n-1}$ .
Arithmetic: $S_{12} = \frac{12}{2}(240 + 11 \times 15) = 6 \times 405 = 2430$ seedlings.
$u_6 = 120 \times 1.08^5 \approx 176.3$ t; $S_6 = 120 \times \frac{1.08^6 - 1}{0.08} \approx 880.3$ t.
$w_{n+1} = 0.9w_n + 2000$ ; $L = \frac{2000}{0.1} = 20\,000$ m³.
$u_{2} = 2.25$ , $u_3 = \frac{1}{2}\left(2.25 + \frac{5}{2.25}\right) \approx 2.236\,11$ ( $\sqrt{5} \approx 2.236\,07$ ). The limit satisfies $L = \frac{1}{2}\left(L + \frac{5}{L}\right)$ , so $2 L^{2} = L^{2} + 5$ , $L^{2} = 5$ and $L = \sqrt{5}$ .

1Where is the sequence going?

2Rules for limits

3Sequences in real life

Where is the sequence going?

Rules for limits

Sequences in real life