Difficulty: Intermediate | Prerequisites: Limits from Calculus I, L'Hopital's Rule, properties of exponentials and logarithms.
Sequences are the gateway to series. Before you can decide whether an infinite sum converges, you need to know whether the individual terms approach zero, and how fast. This section covers the mechanics of evaluating limits of sequences, including rational sequences (polynomial over polynomial), geometric-type sequences, and sequences that require logarithmic tricks. If you are solid on limits from Calc I, this is largely the same toolkit applied to discrete indexing.
TL;DR
A sequence converges if its terms approach a finite limit as n goes to infinity; otherwise it diverges. For rational sequences, divide through by the highest power of n. For exponential-type limits like (1 + a/n)^n, take the natural log, apply L'Hopital's Rule, and exponentiate back.
Sequence
An ordered list of numbers a₁, a₂, a₃, ... defined by a rule aₙ = f(n). Think of it as a function whose input is restricted to positive integers.
Convergence of a sequence
A sequence {aₙ} converges if lim(n→∞) aₙ = L for some finite number L. In simple terms, the terms settle down to a single value and stay arbitrarily close to it.
Divergence of a sequence
A sequence diverges if it does not converge: either the terms grow without bound, or they oscillate without settling.
Geometric sequence
A sequence of the form aₙ = a·rⁿ, where r is the common ratio. Converges to 0 when |r| < 1, diverges when |r| ≥ 1 (except r = 1 gives constant a).
L'Hopital's Rule (for sequences)
If lim(n→∞) aₙ produces an indeterminate form (0/0 or ∞/∞), you can treat n as a continuous variable x and apply L'Hopital's Rule to lim(x→∞) f(x)/g(x). If the continuous limit exists, the sequence limit equals it.
When aₙ is a ratio of two polynomials in n, divide every term in the numerator and denominator by the highest power of n that appears.
Worked example (Exam Problem 4): Find the limit of aₙ = (5 − 9n + 3n⁴) / (4n⁴ − 9n³ + 8).
The highest power is n⁴. Divide numerator and denominator by n⁴.
Numerator becomes: 5/n⁴ − 9/n³ + 3
Denominator becomes: 4 − 9/n + 8/n⁴
As n → ∞, all the 1/nᵏ terms vanish, leaving 3/4.
The sequence converges to 3/4.
The rule of thumb: if the highest power is the same top and bottom, the limit is the ratio of the leading coefficients. If the top has a higher power, the sequence diverges. If the bottom has a higher power, the limit is 0.
For a sequence aₙ = c + rⁿ where |r| < 1, the rⁿ term vanishes as n → ∞.
Worked example (Exam Problem 5): Find the limit of aₙ = 5 + (0.3)ⁿ.
Since |0.3| < 1, lim(n→∞) (0.3)ⁿ = 0.
Therefore lim(n→∞) aₙ = 5 + 0 = 5.
When a sequence has the form (1 + a/n)ⁿ, direct substitution gives the indeterminate form 1^∞. The standard approach is to take the natural log, convert the exponent into a quotient, and apply L'Hopital's Rule.
Worked example (Exam Problem 6): Find the limit of aₙ = (1 + 3/n)ⁿ.
Let L = lim(n→∞) (1 + 3/n)ⁿ. Take ln of both sides: ln L = lim(n→∞) n·ln(1 + 3/n).
Rewrite as: ln L = lim(n→∞) ln(1 + 3/n) / (1/n). This is a 0/0 indeterminate form.
Apply L'Hopital's Rule: differentiate top and bottom with respect to n.
Numerator derivative: (1/(1 + 3/n))·(−3/n²) = −3/(n²(1 + 3/n))
Denominator derivative: −1/n²
The ratio simplifies to: 3/(1 + 3/n), which approaches 3 as n → ∞.
So ln L = 3, meaning L = e³.
Limit of a rational sequence (leading-term rule):
If aₙ = (cₚnᵖ + ...) / (dqnq + ...), then lim(n→∞) aₙ = cₚ/dq when p = q, ∞ (diverges) when p > q, and 0 when p < q.
Geometric sequence limit:
lim(n→∞) rⁿ = 0 if |r| < 1, diverges if |r| > 1, equals 1 if r = 1, does not exist if r = −1.
Compound-growth limit:
lim(n→∞) (1 + a/n)ⁿ = eᵃ
The limit (1 + r/n)ⁿ → eʳ is the foundation of continuous compound interest in finance. If a bank compounds interest n times per year at annual rate r, the effective multiplier as n grows large is exactly eʳ. Sequence convergence also underpins iterative algorithms in computing, where you need to know that successive approximations are actually approaching a solution.
Students often confuse the limit of a sequence with the limit of a series. The sequence {aₙ} having limit 0 does not mean the series Σaₙ converges. Limit zero is necessary but not sufficient for series convergence.
When dealing with (1 + a/n)ⁿ, students sometimes try to argue that since 1 + a/n → 1, the whole thing must equal 1. The exponent n is growing at the same time the base approaches 1, producing the indeterminate form 1^∞.
In rational sequences, students occasionally divide by the wrong power of n. Always divide by the highest power that appears in the entire expression (numerator or denominator), not just the numerator.
⚠️ The "divide by the highest power" method for rational sequences is worth practising until it is automatic. It appears on nearly every exam.
⚠️ The (1 + a/n)ⁿ → eᵃ limit is a standard exam problem. Know the log trick and be comfortable with L'Hopital's Rule in this context.
⚠️ Recognising geometric sequences (|r| < 1 converges, |r| ≥ 1 diverges) is a building block for geometric series, which come next.
True or false: If lim(n→∞) aₙ = 0, then the series Σaₙ must converge.
False. The harmonic series Σ1/n diverges even though 1/n → 0.
Fill in the blank: lim(n→∞) (1 + 5/n)ⁿ = ______.
e⁵.
True or false: The sequence aₙ = (−1)ⁿ converges.
False. It oscillates between −1 and 1 without settling.
Fill in the blank: For aₙ = (7n³ + 2) / (3n³ − n), lim(n→∞) aₙ = ______.
7/3 (ratio of leading coefficients, same degree top and bottom).
Q: Find the limit of the sequence aₙ = (2n² + 5n) / (7n² − 3), or state that it diverges.
A: Divide by n². Numerator: 2 + 5/n. Denominator: 7 − 3/n². As n → ∞, this gives 2/7. The sequence converges to 2/7.
Q: Does the sequence aₙ = n² / eⁿ converge? If so, find its limit.
A: Yes. As n → ∞, exponential growth dominates polynomial growth. By L'Hopital's Rule (applied twice, treating n as continuous x): lim(x→∞) x²/eˣ = lim 2x/eˣ = lim 2/eˣ = 0. The sequence converges to 0.
Q: Evaluate lim(n→∞) (1 + 2/n)^(3n).
A: Rewrite as [(1 + 2/n)ⁿ]³. Since lim(n→∞) (1 + 2/n)ⁿ = e², the answer is (e²)³ = e⁶.
Q: Does the sequence aₙ = (−1)ⁿ · n / (n + 1) converge?
A: No. The absolute value |aₙ| = n/(n+1) → 1, but the sign alternates. The terms approach +1 and −1 alternately, so the sequence diverges.
Sequence limits are the foundation for everything in the series chapter that follows. The Divergence Test for series (if aₙ does not approach 0, the series diverges) is a direct application of sequence limits.
The (1 + a/n)ⁿ → eᵃ result connects to the definition of the number e itself and reappears in differential equations (continuous growth models) and probability (Poisson distribution as a limit of binomials).
Related Terms / Search Tags
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