8.01c–8.01e - Sequence behaviour and Fibonacci numbers | OCR A-Level Further Mathematics A

Describe how a sequence behaves and decide what a claimed limit actually means. In 8.01c–8.01d you will distinguish convergence, divergence, oscillation, periodicity and monotonicity, including sequences of differences or ratios; in 8.01e you will explore Fibonacci and Lucas numbers and the golden ratio.

Describing sequence behaviour

The ratios $1, 2, \frac32, \frac53, \frac85, \frac{13}{8}, \ldots$ bounce above and below a number near $1.618$ . Can a sequence keep changing direction and still have a limit, and why does this particular number appear?

Each fraction divides one Fibonacci number by the preceding one. You will identify their limit later; first separate the words that describe the journey from the word that describes its destination.

You already know

A recurrence system specifies its initial values and index range as well as its update rule. A closed form lets you study $u_n$ directly; for example, $1/n$ tends to $0$ as positive integers $n$ increase.

The terms $1, \frac12, \frac13, \ldots$ decrease towards $0$ . The terms $-1, \frac12, -\frac13, \frac14, \ldots$ change sign and also approach $0$ . Changing direction does not prevent convergence.

Convergence and divergence

A sequence converges to a finite limit $L$ if its terms approach $L$ as $n\to\infty$ , written $u_n\to L$ or $\lim_{n\to\infty}u_n=L$ . A sequence without a finite limit diverges. For example, $1/n$ converges to $0$ , while $n$ diverges to infinity and $(-1)^n$ diverges despite being bounded.

“Bounded” means trapped between fixed finite numbers. It is useful evidence, but boundedness alone does not produce a single destination.

Monotonic, periodic and oscillating

A sequence is monotonically increasing if $u_{n+1}\ge u_n$ at every permitted index, and monotonically decreasing if $u_{n+1}\le u_n$ ; use “strictly” when every inequality is strict. A periodic sequence repeats after a fixed positive integer $p$ : $u_{n+p}=u_n$ . Its period is the least such $p$ . An oscillating sequence repeatedly changes direction or alternates about a value. For example, $(-1)^n$ has period $2$ , whereas $(-1)^n/n$ oscillates but is not periodic.

Some descriptions can apply together. A constant sequence is monotonic in both non-strict senses, periodic with period $1$ , and convergent. A non-constant periodic sequence cannot converge: each distinct value returns indefinitely. Oscillating sequences can converge, remain bounded without converging, or become unbounded.

[DIAGRAM: asset_name: Four different long-term behaviours; asset_slug: h245_8_01_sequence_behaviour; caption: Oscillation can accompany convergence or divergence; boundedness alone does not imply convergence; description: Four panels plot the discrete terms 1 over n, (-1) to the n over n, (-1) to the n and (-1) to the n times n for positive indices. Panels identify monotonic convergence to zero, oscillating convergence to zero, bounded periodic divergence of period two, and unbounded oscillating divergence. Only discrete points are drawn.]
Oscillation can accompany convergence or divergence

Use each word for a separate feature. Calling $(-1)^n$ “oscillating” does not say whether it converges; calling it “bounded and divergent with period $2$ ” settles both questions.

Key idea

A limit concerns one eventual destination. Monotonicity, periodicity and oscillation concern the pattern of the terms.

The following claim confuses a shrinking oscillation with a repeated cycle.

Monotonicity and limits from formulae

A sign calculation proves monotonicity for every index, whereas a list of decimal values only suggests it.

For a real sequence, inspect $u_{n+1}-u_n$ . A positive difference proves strict increase; a negative one proves strict decrease. For a positive sequence, comparing $u_{n+1}/u_n$ with $1$ is another option. Do not divide an inequality by a term whose sign is unknown.

Successive differences

For every index in the sequence's domain,

u_{n+1}-u_n>0\ \Rightarrow\ \text{strict increase},\qquad u_{n+1}-u_n<0\ \Rightarrow\ \text{strict decrease}

Non-strict inequalities give the corresponding non-strict monotonicity.
Not in the formula booklet: learn it.

A rational closed form often reveals its limit by dividing numerator and denominator by the highest power of $n$ , or by rewriting it as a constant plus a vanishing term.

Worked example 1: approaching a limit from below

Question. For $u_n=(3n+1)/(n+2)$ , $n\ge0$ , show that the sequence is strictly increasing and find its limit.

Plan. Rewrite the formula to show both the changing part and the limiting constant.

\begin{aligned} u_n&=3-\frac5{n+2} && \text{divide the numerator} \\ u_{n+1}-u_n&=\frac5{n+2}-\frac5{n+3} && \text{subtract successive terms} \\ &=\frac5{(n+2)(n+3)}>0 && n\ge0 \\ \lim_{n\to\infty}u_n&=3 && \frac5{n+2}\to0 \end{aligned}

Why this step? The rewrite identifies a positive quantity being subtracted from $3$ , and that quantity shrinks as the index grows.

Answer. The sequence is strictly increasing and converges to $3$ , from below.

Check. $u_0=\frac12$ , $u_1=\frac43$ , $u_2=\frac74$ ; all are below $3$ and increase.

The next example changes only the constant term in the numerator. That reverses the side from which the same limit is approached.

Worked example 2: approaching the same limit from above

Question. For $v_n=(3n+11)/(n+2)$ , $n\ge0$ , show that the sequence is strictly decreasing and find its limit.

Plan. Use the same rewrite and difference calculation, keeping the changed sign.

\begin{aligned} v_n&=3+\frac5{n+2} && \text{divide the numerator} \\ v_{n+1}-v_n&=\frac5{n+3}-\frac5{n+2} && \text{subtract successive terms} \\ &=-\frac5{(n+2)(n+3)}<0 && n\ge0 \\ \lim_{n\to\infty}v_n&=3 && \frac5{n+2}\to0 \end{aligned}

Why this step? The same positive vanishing term is now added to $3$ , so the terms lie above $3$ and decrease.

Answer. The sequence is strictly decreasing and converges to $3$ , from above.

Check. $v_0=\frac{11}{2}$ , $v_1=\frac{14}{3}$ , $v_2=\frac{17}{4}$ , decreasing towards $3$ .

These rational-sequence results are not in the formula booklet: derive them. A sequence may become monotonic only after some index; state the range on which your sign argument works rather than describing the entire sequence from a late pattern.

Watch out

Wrong: “The first three differences are positive, so the sequence is increasing.” Right: show $u_{n+1}-u_n>0$ for every permitted $n$ . A finite list cannot exclude a later reversal.

The next formula changes the coefficient of $n$ as well as its limiting value; the algebraic test stays the same.

Steady states and derived sequences

If a recurrence sequence converges, its consecutive terms have the same limit. Substituting that limit finds possible destinations; it does not prove that the sequence reaches one.

For example, $u_{n+1}=2u_n$ has the fixed point $0$ . Starting at $u_0=1$ gives $1, 2, 4, \ldots$ , so solving $L=2L$ has not established convergence.

Steady state

A steady state is a value which remains unchanged by the update. For $u_{n+1}=f(u_n)$ , it satisfies $L=f(L)$ . Starting at that value produces a constant sequence. If a sequence governed by a continuous update $f$ has a finite limit $L$ , that limit must satisfy the same equation.

To establish a limit, use a given closed form, a justified bound and monotonicity, or other information in the question. A monotonic bounded real sequence converges; its bound need not itself be its limit. This convergence principle is not in the formula booklet: learn it.

Worked example 3: proving a candidate limit is reached

Question. A sequence satisfies $u_{n+1}=\frac12u_n+3$ , $u_0=4$ , and has closed form $u_n=6-2(\frac12)^n$ . Find the steady state and the limits of $u_n$ and $d_n=u_{n+1}-u_n$ .

Plan. Find the fixed point, then use the supplied closed form to justify the limits.

\begin{aligned} L&=\tfrac12L+3\quad\Rightarrow\quad L=6 && \text{steady-state equation} \\ u_n&\to6 && (\tfrac12)^n\to0 \\ d_n&=\left[6-2(\tfrac12)^{n+1}\right]-\left[6-2(\tfrac12)^n\right] && \text{form differences} \\ &=(\tfrac12)^n\to0 && \text{combine powers} \end{aligned}

Why this step? The closed form, rather than the fixed-point equation alone, proves convergence from the given initial value.

Answer. The steady state and the limit of $u_n$ are $6$ ; the difference sequence tends to $0$ .

Check. $u_0=4$ , $u_1=5$ , $u_2=\frac{11}{2}$ , giving differences $1, \frac12, \frac14$ , as predicted.

The next example changes only the sign of the multiplier. It preserves convergence while making the differences alternate in sign.

Worked example 4: an oscillating approach to a steady state

Question. A sequence satisfies $v_{n+1}=-\frac12v_n+3$ , $v_0=4$ , and has closed form $v_n=2+2(-\frac12)^n$ . Find its steady state and the limits of $v_n$ , $d_n=v_{n+1}-v_n$ , and $q_n=d_{n+1}/d_n$ .

Plan. Solve for the fixed point, then subtract and divide successive difference formulae.

\begin{aligned} L&=-\tfrac12L+3\quad\Rightarrow\quad L=2 && \text{fixed point} \\ v_n&\to2 && (-\tfrac12)^n\to0 \\ d_n&=2(-\tfrac12)^n(-\tfrac12-1)=-3(-\tfrac12)^n && \text{factor the difference} \\ d_n&\to0,\qquad q_n=-\tfrac12 && \text{take limits and ratios} \end{aligned}

Why this step? The difference is a sequence in its own right. Dividing its consecutive terms cancels its non-zero coefficient and power.

Answer. The sequence oscillates about the steady state $2$ and converges to it. The difference sequence tends to $0$ , while $q_n\to-\frac12$ .

Check. $v_0=4, v_1=1, v_2=\frac52, v_3=\frac74$ , giving differences $-3, \frac32, -\frac34$ ; their ratios equal $-\frac12$ .

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8.01c–8.01e - Sequence behaviour and Fibonacci numbers

Describing sequence behaviour

Monotonicity and limits from formulae

Steady states and derived sequences

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