Limit of a Sequence

Video lesson · from 10:502 worked examples

One lesson video covers all of Recurrence Relations, so it opens at 10:50 for this topic — not from the beginning.

If we have a recurrence relation of the form $U_{n+1} = aU_n + b$

where $a < -1 \text{ or } a > 1$ then the sequence will diverge.

i.e. it will keep increasing, decreasing or oscillating.

If we have a recurrence relation of the form $U_{n+1} = aU_n + b$

where $-1 < a < 1$ then the sequence will converge to a limit.

i.e. it will settle at approximately the same value.

For convergent sequences defined by $U_{n+1} = aU_n + b$ with $-1 < a < 1$ , $U_n$ tends to a limit $L$ as $n \to \infty$ .

L = aL + b

L - aL = b

L(1 - a) = b

L = \frac{b}{1 - a}

A limit exists only when $-1 \lt a \lt 1$ : state this condition. Quoting a limit without justifying that it exists loses marks.
$L = \frac{b}{1-a}$ : note it is $1-a$ on the bottom, not $a-1$ . Getting this the wrong way round flips the sign.
The limit does not depend on the starting value: $u_0$ affects how fast you get there, not where you end up.
Interpret it in context: say what the limit means in the situation described — a long-term level, not just a number.

Example 1

A sequence is defined by $U_n = 0.6U_{n-1} + 5$ with $U_1 = 10$ .

a) Does the sequence have a limit as $n \to \infty$ ?

b) Calculate this limit if it exists.

a) Yes, a limit exists because $-1 < 0.6 < 1$ .

b) Using the formula $L = \frac{b}{1 - a}$ :

\begin{aligned} L &= \frac{5}{1 - 0.6} \\ &= \frac{5}{0.4} \\ &= 12.5 \end{aligned}

The limit is 12.5.

Example 2

The deer population in a forest is estimated to drop by 7.3% each year.

Each year, 20 deer are introduced to the forest. The initial deer population is 200.

a) How many deer will there be in the forest after 3 years?

b) What is the long-term effect on the population?

a) Dropping by 7.3% means multiplying by $1 - 0.073 = 0.927$ . Adding 20 each year.

U_{n+1} = 0.927U_n + 20, \quad U_0 = 200

U_1 = 0.927(200) + 20 = 205.4

U_2 = 0.927(205.4) + 20 = 210.4058

U_3 = 0.927(210.4058) + 20 \approx 215.046

There will be approximately 215 deer after 3 years.

b) To find the long-term effect, calculate the limit:

Limit exists since $-1 < 0.927 < 1$ .

\begin{aligned} L &= \frac{b}{1 - a} \\ &= \frac{20}{1 - 0.927} \\ &= \frac{20}{0.073} \\ &\approx 273.97 \end{aligned}

The population will settle at around 274 deer.