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Subsequence

Definition

Where this appears in the course

Subsequences are not defined in the 2026-27 lectures. The definition is given inside the tutorial question that uses it (Tutorial 2), and this page records it in the same form.

Let (an)n≥m(a_n)_{n\ge m} be a sequence. A subsequence of (an)(a_n) is a new sequence (ank)k≥1(a_{n_k})_{k\ge 1} obtained by selecting terms of (an)(a_n) along a strictly increasing sequence of indices (nk)k≥1(n_k)_{k\ge 1} with nk≥mn_k \ge m:

(ank)k≥1=an1, an2, an3, …,(a_{n_k})_{k\ge 1} = a_{n_1},\,a_{n_2},\,a_{n_3},\,\dots,

where m≤n1<n2<n3<⋯ .m \le n_1 < n_2 < n_3 < \cdots.

Interpretation

A subsequence is what you get when you thin out a sequence while keeping the original order. You choose infinitely many of its terms, further and further along, and you may not reorder or repeat them.

The terms (-1) to the n times (1 + 1/n), alternating around 1 and -1, with the even-indexed terms highlighted as a subsequence approaching 1 and the odd terms skipped.

The figure keeps the even-indexed terms of an=(−1)n(1+1n)a_n=(-1)^n\left(1+\frac{1}{n}\right), so nk=2kn_k=2k and ank=1+12ka_{n_k}=1+\frac{1}{2k}. These terms, a2=1.5a_2=1.5, a4=1.25a_4=1.25, a6≈1.17,…a_6\approx1.17,\dots, decrease to 11, while the skipped odd-indexed terms tend to −1-1.

Choose the indices nkn_k: the even and odd subsequences of an=(−1)n(1+1n)a_n=(-1)^n\left(1+\frac1n\right) converge to different limits, which is why the whole sequence diverges.

The condition n1<n2<⋯n_1<n_2<\cdots makes a subsequence a sequence in its own right. Because the indices are integers that strictly increase, nk≥kn_k\ge k when the sequence starts at n=1n=1, so the subsequence runs through the tail of (an)(a_n) as kk grows. This is why every subsequence of a sequence with limit LL also has limit LL. Lists such as a1,a1,a2,a3,…a_1,a_1,a_2,a_3,\dots or a2,a1,a4,a3,…a_2,a_1,a_4,a_3,\dots are not subsequences.

Subsequences are the usual tool for proving divergence: two subsequences with different limits, as in the figure, show that (an)(a_n) has no limit. They also return in Math III, where the Bolzano-Weierstrass theorem states that every bounded sequence has a convergent subsequence.

A common slip is to run the implication backwards. One convergent subsequence says nothing about the whole sequence: (−1)n(-1)^n has the constant subsequence 1,1,1,…1,1,1,\dots but diverges.

Key Results

  • Every subsequence of a sequence uses terms from the original sequence, but with a new index set n1<n2<⋯.n_1<n_2<\cdots\text{.}
  • If an→L,a_n \to L\text{,} then every subsequence anka_{n_k} also converges to L.L\text{.}
  • Subsequence limits help detect non-convergence: if two subsequences converge to different limits, then the original sequence does not converge.

Examples

  • If an=(−1)n,a_n = (-1)^n\text{,} then (a2k)k≥1=1,1,1,…(a_{2k})_{k\ge 1} = 1,1,1,\dots and (a2k−1)k≥1=−1,−1,−1,…(a_{2k-1})_{k\ge 1} = -1,-1,-1,\dots are subsequences.
  • If an=1n,a_n = \dfrac{1}{n}\text{,} then a2k=12ka_{2k}=\dfrac{1}{2k} is a subsequence, and it still converges to 0.0\text{.}

Prerequisites

Sources

Last updated: 20 October 2026 03:02