A sequence is an endless line of numbered terms.
If the terms get as close as you like to one value as the index n grows, that value is the limit, and the sequence converges to it.
If the terms do not gather at one value, growing without bound or swinging between two values, the sequence diverges.
In the pictures, increasing n brings aₙ = 1 + 1/n toward L = 1, and shrinking ε pushes back the index N at which the terms start staying inside the band.
The Value a Sequence Approaches
A sequence has no last term, because its index n keeps growing forever. So the limit of a sequence is not a question about any particular term; it asks whether the terms, once the index is large enough, stay near a single value L. If, however narrow a band you draw around L, there is always an index after which every term lies inside it, the sequence converges to L; if no such L exists, it diverges. A limit of a function deals with a real number x approaching a, while a limit of a sequence deals only with the natural number n growing without bound. With this definition in place, the limit of a sum, difference or product of two convergent sequences is the sum, difference or product of their limits, and the same goes for a quotient as long as the limit of the denominator is not 0.
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The first 20 terms of aₙ = 1 + 1/n and the limit L = 1
The dots in the first picture are the first 20 terms of aₙ = 1 + 1/n, and the red dashed line is the limit L = 1. Terms up to the n-th one chosen with the slider light up in gold, and the top band shows that term's value and its distance from the limit line, |aₙ − 1| = 1/n. As n goes from 1 to 20 the distance shrinks through 1, 1/2, 1/3, … down to 0.05; the steps get smaller and smaller, but the distance never reaches 0. In the second picture the ε slider sets the width of the green band around L. Terms inside the band are green, terms outside are gold, and the white dashed line marks N, the first index from which every term is inside. Making ε smaller pushes N further right, but whatever ε you choose, such an N always exists.
5
👀 What to read in the picture
①The gold dots are the terms and the red dashed line is the limit L = 1
②As n grows, the distance 1/n between a term and the line shrinks toward 0
③No term reaches distance 0, but whatever small distance you pick, every term from some index on is closer than that. That is convergence
Saying It Exactly with ε and N
Terms that keep growing do not mean the sequence must diverge. In the third picture, n/(n + 1) only ever increases, yet it never passes 1 and converges to 1. On the other hand, (−1)ⁿ(1 + 1/n) has terms whose size approaches 1, but the sign keeps flipping, so the terms alternate between near 1 and near −1 and never gather at one value; the sequence diverges. Divergence to infinity and divergence by oscillation are both divergence, but they look different, and it helps to tell them apart. Nor can you judge from the first few terms. Convergence is a statement about all terms after some index, so however far away the early terms are, the conclusion does not change. For rational expressions, divide by the highest power and see whether terms like 1/n go to 0; for a bounded factor such as sin n, squeeze it from above and below.
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The band |aₙ − 1| < ε around L and the index N from which every term is inside
0.1
📏 The ε band and the index N
①ε is the error allowed around the limit, that is, the half-width of the band
②For aₙ = 1 + 1/n a term is inside the band when 1/n < ε, so you can take N = ⌊1/ε⌋ + 1. With ε = 0.1, N = 11
③That such an N exists however small ε is chosen is the exact definition of convergence
ε-N definition
∀ε > 0, ∃N : n ≥ N ⇒ |an − L| < ε
For every positive ε there is an index N such that every term with n ≥ N satisfies |aₙ − L| < ε
Convergence and Divergence Compared
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The first 20 terms of three sequences; dashed lines mark the heights the terms approach
1+1/n
🔍 Three behaviors
①1 + 1/n approaches 1 from above and converges
②(−1)ⁿ(1 + 1/n) alternates between near 1 and near −1, never gathering at one value, so it diverges (oscillates)
③n/(n + 1) approaches 1 from below and converges. A sequence can keep increasing and still converge
Basic Properties of Limits
Limits of sums, differences and products
lim(an ± bn) = α ± β, lim(an · bn) = α · β
If lim aₙ = α and lim bₙ = β, the limit of a sum, difference or product is the sum, difference or product of the limits
Limit of a quotient
lim anbn = αβ (β ≠ 0)
If β ≠ 0 and bₙ ≠ 0, the limit of the quotient is the quotient of the limits
Squeeze theorem
an ≤ cn ≤ bn, lim an = lim bn = L ⇒ lim cn = L
A sequence trapped between two sequences that close in on the same value converges to that value too
Worked Examples
Example 1
Find limn→∞2n+3n-1.
1
Divide the numerator and the denominator by n.
2n+3n-1 = 2 + 3/n1 - 1/n
2
As n → ∞, 3/n → 0 and 1/n → 0.
2 + 01 - 0 = 2
▸ 2
For a limit of a rational expression, the basic move is to divide the numerator and denominator by the highest power in the denominator.
Example 2
Find limn→∞sin nn.
1
Since −1 ≤ sin n ≤ 1, divide every part by the positive number n.
-1n ≤ sin nn ≤ 1n
2
Both outer limits are 0, so by the squeeze theorem the middle one is 0 too.
limn→∞ (-1/n) = limn→∞1n = 0
▸ 0
A sequence with bounded values, such as sin or cos, can be squeezed between ±1/n.
Summary
Limits used most
limn→∞1np = 0 (p > 0), limn→∞ rn = 0 (|r| < 1)
For p > 0, 1/nᵖ converges to 0; for |r| < 1, rⁿ converges to 0
CSAT-style problem, adapted
What is limn→∞ (√(n2+3n) − n)?
①1
②32
③2
④3
⑤∞ (diverges)
▸ ② 32
1
It has the form ∞ − ∞ with a root, so multiply by the conjugate to rationalize.
√(n2+3n) − n = 3n√(n2+3n) + n
2
Dividing the numerator and denominator by n makes the denominator converge to 2.
3√(1 + 3/n) + 1 → 32
🎯 Exam Points
①Convergence means every term after some index lies in a band around the limit
②Sums, differences, products and quotients of convergent sequences are computed from their limits
③An oscillating sequence has no limit, so it diverges
④Squeeze a bounded factor from above and below and use the squeeze theorem
⑤Divide a rational expression by the highest power in the denominator and use 1/nᵖ → 0