01 · A decreasing rule
Show that uₙ = 2 + 1/n is strictly decreasing.
Hint
Subtract consecutive terms.
Worked solution
uₙ₊₁ − uₙ = −1/[n(n + 1)] < 0 for every n ≥ 1.
Understand · explore · practise
Increasing, decreasing and periodic sequences
Prove when a sequence increases or decreases, identify its least period, calculate periodic sums and analyse recurrence behaviour with original practice.
Before you startSequence formulas, recurrence relations and inequalities
01 / What increasing means
The condition must hold at every relevant index.
Strictly increasing: uₙ₊₁ > uₙ for every n
Strictly decreasing: uₙ₊₁ < uₙ for every n
Here “increasing” and “decreasing” mean strict inequalities. “Nondecreasing” allows equality, while “nonincreasing” allows equality in the opposite direction. A constant sequence is neither strictly increasing nor strictly decreasing.
A list can rise for a while and later fall. A few calculated terms may suggest a pattern, but a general argument is needed to establish behaviour for all n.
02 / Prove it using consecutive differences
Find the sign of uₙ₊₁ − uₙ.
uₙ = n/(n + 2), for n ≥ 1Worked example
uₙ₊₁ − uₙ = (n + 1)/(n + 3) − n/(n + 2)
Substitute n + 1 into the entire rule.
= [(n + 1)(n + 2) − n(n + 3)]/[(n + 2)(n + 3)]
Use a common denominator.
= 2/[(n + 2)(n + 3)] > 0
Both denominator factors are positive, so the sequence is strictly increasing.
vₙ = 4 + 5/nWorked example
vₙ₊₁ − vₙ = 5/(n + 1) − 5/n
The constant 4 cancels.
= −5/[n(n + 1)] < 0 for n ≥ 1
Therefore vₙ is strictly decreasing.
Comparing a ratio with 1 only works with appropriate sign conditions. The difference method avoids dividing an inequality by a possibly negative term.
03 / Arithmetic and geometric cases
The first-term sign matters for geometric sequences.
For an arithmetic sequence, uₙ₊₁ − uₙ = d. Thus d > 0 gives increasing terms, d < 0 gives decreasing terms, and d = 0 gives a constant sequence.
For a geometric sequence with r > 0, the difference is ar^(n − 1)(r − 1). Its sign depends on a(r − 1), not just r.
Negative first termsWorked example
−4, −8, −16,… has r = 2
It decreases because the terms become more negative.
−4, −2, −1,… has r = ½
It increases towards zero despite the ratio being below 1.
With a ≠ 0 and r < 0, signs alternate, so the infinite sequence is neither increasing nor decreasing. With r = 0, later terms are all zero: the sequence is not strictly monotone. With a = 0 or r = 1, it is constant.
04 / Parameter conditions for all n
Identify the smallest consecutive difference.
For which real k is uₙ = n² − kn strictly increasing for all n ≥ 1?Worked example
uₙ₊₁ − uₙ = 2n + 1 − k
Expand the difference before considering its sign.
The smallest difference occurs at n = 1 and is 3 − k
All later differences are larger.
Require 3 − k > 0, giving k < 3
This condition is both necessary and sufficient.
At k = 3 the first two terms are equal, while later terms rise: the sequence is nondecreasing but not strictly increasing. For k > 3 the first step decreases, but sufficiently late steps increase, so the whole sequence is not monotone.
05 / Behaviour from a recurrence
First establish a bound that remains true at the next step.
u₁ = 2 and uₙ₊₁ = ½uₙ + 3Worked example
u₁ < 6
The sequence starts below 6.
If uₙ < 6, then uₙ₊₁ = ½uₙ + 3 < 6
Every step preserves that bound, so all terms remain below 6.
uₙ₊₁ − uₙ = 3 − ½uₙ = (6 − uₙ)/2 > 0
The preserved bound makes every difference positive.
Therefore the sequence is strictly increasing. Simply observing 2, 4, 5, 5.5 is weaker than this all-step argument. Starting above 6 would instead give decreasing terms; starting at 6 would give a constant sequence.
06 / What periodic means
A whole pattern repeats after a fixed positive integer shift.
p is a period if uₙ₊ₚ = uₙ for every n.
The least period is the smallest such positive integer p.
The repeating sequence 4, −2, 1, 4, −2, 1,… has least period 3. Six is also a period, but it is not the least. A constant sequence has least period 1.
Alternating signs do not guarantee periodicity: 1, −2, 3, −4,… never returns to the same magnitude at a later index. A nonconstant periodic sequence cannot be strictly increasing or decreasing across all indices, because it returns to earlier values.
Track a term and its running totalExplore
Dots show individual terms. The highlighted term and running sum use the index you choose. The vertical scale adjusts to the chosen rule.
u₈ = −2; S₈ = 8.
2 complete cycles contribute 6; 2 leftover terms contribute 2.
Watch complete cycles and leftover terms
Pause, replay or seek freely. The notes explain the same idea and stay in view.
07 / Distant terms and periodic sums
Split the index into complete cycles and a remainder.
Repeat the cycle 3, −1, 2Worked example
100 = 33 × 3 + 1
Term 100 is in the first position of a cycle, so u₁₀₀ = 3.
Each cycle sums to 3 − 1 + 2 = 4
A periodic sequence need not have zero cycle sum.
23 = 7 × 3 + 2
The first 23 terms contain seven whole cycles and two extra terms.
S₂₃ = 7 × 4 + (3 − 1) = 30
Include the initial part of the next cycle in order.
A zero remainder puts a term at the last position of a cycle: for period 3, term 99 is the third position, not a nonexistent position 0. A sum with zero remainder has only whole cycles and no extra terms.
Periodicity of the terms does not make their partial sums periodic when the cycle sum is nonzero. For this example each complete cycle adds another 4.
08 / A periodic recurrence extension
For a two-step rule, a repeated pair restarts the same process.
u₁ = 2, u₂ = 3 and uₙ₊₂ = (1 + uₙ₊₁)/uₙWorked example
u₃ = 2, u₄ = 1, u₅ = 1
Each denominator is positive.
u₆ = 2 and u₇ = 3
The pair (u₆, u₇) equals the original pair (u₁, u₂).
The five-term block 2, 3, 2, 1, 1 repeats
The same pair of inputs produces the same next pair.
Its least period is 5
Shifts 1, 3 and 4 fail at the first term; shift 2 fails at the second term.
A single repeated value is not enough for a two-step rule: u₁ = u₃ here, but the sequence does not have period 2.
Why positive starting values always give a five-step repeatWorked example
u₁ = a > 0, u₂ = b > 0
The positive starting values make each subsequent fraction positive.
u₃ = (1 + b)/a
Apply the rule once.
u₄ = (a + b + 1)/(ab)
Substitute u₃ and simplify.
u₅ = (a + 1)/b
The factor 1 + b cancels, and is positive.
u₆ = a, u₇ = b
The original pair returns, establishing that 5 is a period.
Five need not be the least period for every starting pair. If a = b = (1 + √5)/2, the sequence is constant and has least period 1. The positivity assumption above is essential to this proof’s denominator checks.
09 / Your turn
Give a reason that applies to the full sequence.
Indices start at n = 1 unless another start is given.
02 · An increasing rule
Show that uₙ = n/(n + 3) is strictly increasing.
Hint
Use a common denominator.
Worked solution
uₙ₊₁ − uₙ = 3/[(n + 3)(n + 4)] > 0 for every n ≥ 1.
03 · A parameter boundary
Find all t for which uₙ = n² + tn is strictly increasing for every n ≥ 1.
Hint
The smallest difference is at n = 1.
Worked solution
The difference is 2n + 1 + t, whose minimum is 3 + t. Thus t > −3. At t = −3 it is only nondecreasing.
04 · A negative first term
Classify uₙ = −7(⅓)^(n − 1).
Hint
Watch the sign of the terms.
Worked solution
It is strictly increasing: the negative terms move towards zero. The difference is (14/3)(⅓)^(n − 1) > 0.
05 · A recurrence bound
u₁ = 10 and uₙ₊₁ = ½uₙ + 3. Show that the sequence decreases.
Hint
Show all terms stay above 6 first.
Worked solution
If uₙ > 6 then uₙ₊₁ − 6 = ½(uₙ − 6) > 0. Since u₁ > 6, this holds at every step. Then uₙ₊₁ − uₙ = (6 − uₙ)/2 < 0.
06 · Alternating cycle
The cycle 7, −7 repeats. Find its least period, u₁₀₁, S₁₀₀ and S₁₀₁.
Hint
Each pair sums to zero.
Worked solution
Least period 2; u₁₀₁ = 7. S₁₀₀ = 0 and S₁₀₁ = 7.
07 · A three-term cycle
Repeat 2, −3, 5. Find u₅₀ and S₅₀.
Hint
50 = 16 × 3 + 2.
Worked solution
u₅₀ = −3. Each cycle sums to 4, so S₅₀ = 16 × 4 + (2 − 3) = 63.
08 · Alternating is not periodic
Is uₙ = (−1)ⁿ/n periodic?
Hint
Compare the magnitudes at different indices.
Worked solution
No. Its magnitudes 1/n strictly decrease, so no later term can equal an earlier term in magnitude. A fixed positive repeating shift is impossible.
09 · A repeated pair
u₁ = 1, u₂ = 2 and uₙ₊₂ = (1 + uₙ₊₁)/uₙ. Find u₃ through u₇ and the least period.
Hint
Compare the pair at indices 6 and 7 with the starting pair.
Worked solution
The terms are 1, 2, 3, 2, 1, 1, 2. The pair repeats after five steps, so 5 is a period. Shifts 1, 2 and 3 fail at the first term; shift 4 fails at the second. The least period is 5.
10 · Constant and periodic
For which b is uₙ = (b − 2)3^(n − 1) constant? What is its least period then?
Hint
Use the first two terms or their difference.
Worked solution
Constancy requires 3(b − 2) = b − 2, giving b = 2. The all-zero sequence has least period 1.
10 / Recap
Prove a global pattern, not just a few matching values.
- Strict monotonicity requires the relevant strict inequality at every step.
- The sign of uₙ₊₁ − uₙ is often the simplest proof.
- Parameters must work for every allowed index.
- Least period means the smallest positive repeating shift.
- Use whole cycles plus leftover terms for distant sums.
- Two-step recurrences require a repeated pair to restart the process.
Section 1 of 10 · What increasing means