Subsequences and Their Limits

The idea of extracting and controlling subsequences reaches far beyond a single lemma about limits. It is the engine behind the Bolzano-Weierstrass Theorem, which guarantees every bounded sequence of reals has a convergent subsequence — the crucial compactness fact underlying existence proofs throughout analysis, from the Extreme Value Theorem to the existence of solutions to differential equations via Peano's existence theorem. The notions of Limit Superior and Limit Inferior are literally defined as the largest and smallest subsequential limits of a bounded sequence, making subsequence theory the direct foundation for handling sequences that oscillate without converging, such as in the ratio and root tests for series. In functional analysis and PDE theory, "weak subsequential limits" generalize this same idea to infinite-dimensional spaces, where compactness arguments extract convergent subsequences from bounded families of functions (Banach–Alaoglu type theorems). In probability theory, tightness arguments and Prokhorov's theorem extract weakly convergent subsequences of probability measures, a direct descendant of the same logical structure. In numerical analysis and optimization, proofs that iterative algorithms (gradient descent, fixed-point iteration) converge often proceed by first showing a subsequence of iterates converges to a critical point, then upgrading this using monotonicity or contraction estimates — mirroring exactly the recursive-sequence strategy of Worked Example 2. In computer science, the study of sequences and limits of discrete structures (e.g. convergent graph sequences in graph limit theory) uses subsequence extraction as its foundational compactness tool.

Ask the tutor: Can you show me why every bounded sequence must have a convergent subsequence, using a bisection argument?

Picture the sequence $(a_n)$ as an infinite list of dots plotted against their index $n$ on a horizontal axis, with the values $a_n$ plotted vertically. A subsequence is what you get by circling infinitely many of these dots — but you are not allowed to circle them out of order. You must circle a dot with a strictly larger index every time you circle a new one. You may skip as many terms as you like between selections (even infinitely many at each stage is fine, as long as each individual index you pick is finite and strictly greater than the last one picked). The key structural fact making the main theorem work is this: the $k$-th selected index $n_k$ is always at least $k$, i.e. $n_k \ge k$ for all $k$. This holds because $n_1 \ge 1$, and each step forces $n_{k+1} \ge n_k + 1$, so by induction $n_k \ge k$. Consequently, once you go far enough out in $k$, the index $n_k$ is guaranteed to be far out in $n$ as well — so if the tail of $(a_n)$ is trapped near $L$, the tail of $(a_{n_k})$ is trapped near $L$ too. Convergence of the parent sequence forces every subsequence into the same net.

Now we prove the main theorem, using the The Epsilon-N Definition of a Limit directly and the index growth lemma just established.

Example 1 — proving divergence of an oscillating, growing sequence. Let

So $a_1 = -3$, $a_2 = 2.5$, $a_3 = -\tfrac{7}{3} \approx -2.333$, $a_4 = 2.25$, $a_5 = -2.2$, and so on. We claim $(a_n)$ diverges, using the corollary above. Consider the subsequence of even indices: $n_k = 2k$, so $n_1 = 2 < n_2 = 4 < n_3 = 6 < \cdots$, a valid strictly increasing index sequence. Then

We compute $\lim_{k\to\infty} a_{2k} = 2$, since $1/(2k) \to 0$ (this itself follows from the Archimedean property applied in the $\varepsilon$-$N$ definition: given $\varepsilon>0$, pick $K > \tfrac{1}{2\varepsilon}$, so $k \ge K \Rightarrow \tfrac{1}{2k} < \varepsilon$). Now consider the subsequence of odd indices: $n_k = 2k - 1$, giving $n_1 = 1 < n_2 = 3 < n_3 = 5 < \cdots$. Then

and $\lim_{k \to \infty} a_{2k-1} = -2$ by an identical argument. We have exhibited two subsequences of $(a_n)$, one converging to $2$ and one converging to $-2$. Since $2 \ne -2$, the corollary tells us $(a_n)$ diverges. Note this is a complete, rigorous proof of divergence — no need to grapple with the oscillating full sequence directly.

Example 2 — subsequences of a sequence built from the Monotone Convergence Theorem. Define $(b_n)$ recursively by

So $b_1 = 1$, $b_2 = \sqrt{5} \approx 2.236$, $b_3 = \sqrt{2\sqrt 5 + 3} \approx \sqrt{7.472} \approx 2.733$, $b_4 \approx \sqrt{8.466} \approx 2.910$, and the terms appear to increase toward something near $3$. Step 1: show $(b_n)$ is increasing and bounded above by $3$, invoking <a class="wiki-link" href="/wiki/math-analysis-monotone-convergence-theorem" data-track="wiki_article.body.wiki_link">The Monotone Convergence Theorem</a>. We show by induction that $1 \le b_n < 3$ for all $n$. Base case: $b_1 = 1$, true. Inductive step: assume $1 \le b_n < 3$. Then $2b_n + 3 \in [5, 9)$, so $b_{n+1} = \sqrt{2b_n+3} \in [\sqrt5, 3)$, in particular $1 \le b_{n+1} < 3$. This closes the induction, so $b_n < 3$ for all $n$. Next show $b_{n+1} > b_n$ for all $n$, i.e. $(b_n)$ is strictly increasing. Since $b_n \ge 1 > 0$ and $b_{n+1} \ge 1 > 0$, the inequality $b_{n+1} > b_n$ is equivalent (after squaring, valid since both sides are positive) to

Since $1 \le b_n < 3$, we have $b_n - 3 < 0$ and $b_n + 1 > 0$, so the product is negative, confirming $b_{n+1} > b_n$. Thus $(b_n)$ is increasing and bounded above, so by the Monotone Convergence Theorem it converges to some limit $L = \sup_n b_n$, with $1 \le L \le 3$. Step 2: identify $L$ using continuity of the algebraic operations, then apply subsequence inheritance. Since $b_n \to L$, the shifted sequence $(b_{n+1})_{n=1}^\infty$ is itself a subsequence of $(b_n)$ — precisely the subsequence with index selection $n_k = k+1$ (strictly increasing: $2 < 3 < 4 < \cdots$). By our theorem, $b_{n+1} \to L$ as well. On the other hand, using the algebraic limit laws (themselves consequences of the $\varepsilon$-$N$ definition) and continuity of $\sqrt{\cdot}$ on $[0,\infty)$,

Since limits are unique, $L = \sqrt{2L+3}$. Squaring (valid, both sides nonnegative), $L^2 = 2L + 3$, i.e. $L^2 - 2L - 3 = 0$, i.e. $(L-3)(L+1) = 0$. So $L = 3$ or $L = -1$. Since $L \ge 1 > 0$, we discard $L = -1$ and conclude

This example shows subsequence inheritance used not merely to detect divergence, but as an active tool: recognizing the shifted sequence $(b_{n+1})$ as a subsequence of $(b_n)$ lets us set up the fixed-point equation that pins down the exact numerical limit $L = 3$.