Convergence in R^n

What It Means for Vectors to Approach a Limit

The previous tutorial introduced sequences in \(\mathbb{R}^n\), their coordinate sequences, and boundedness. Convergence asks a different question: do the vectors eventually lie as close as we require to one fixed vector? The distance must become small for every sufficiently late term, not merely for selected terms. We begin with the Euclidean norm, using the notation \(x^{(k)}\) for the \(k\)-th vector and \(a\) for a proposed limit.

The quantity \(\|x^{(k)}-a\|_2\) is the Euclidean distance from the \(k\)-th term to \(a\). Thus convergence says that, however small a radius \(\varepsilon\) is chosen, all terms from some index onward lie in the open ball \(B_\varepsilon^{(2)}(a)\). The starting index may depend on \(\varepsilon\). There is no requirement that the first \(K-1\) terms lie near \(a\).

The definition is a statement about the tail of the sequence. Changing, deleting, or adding finitely many initial terms does not change whether the sequence converges or what its limit is: for any chosen tolerance, one can move the starting index beyond all the altered terms. By contrast, being close at a few selected indices is not enough. The condition must hold at every index \(k\geq K\).

Basic Consequences of the Definition

A useful first consequence is that a convergent sequence cannot have terms that keep making arbitrarily large excursions. The terms sufficiently far along are close to the limit, while there are only finitely many earlier terms. This turns convergence into a direct source of boundedness.

Proof. Suppose \(x^{(k)}\to a\). Apply the definition with \(\varepsilon=1\). There is a positive integer \(K\) such that, for every \(k\geq K\), \(\|x^{(k)}-a\|_2<1\). The triangle inequality then gives \[ \|x^{(k)}\|_2 \leq \|x^{(k)}-a\|_2+\|a\|_2 <1+\|a\|_2 \qquad (k\geq K). \] The terms with \(k<K\) form a finite set. If there are any such terms, their norms have a finite maximum; if there are none, no additional bound is needed. Taking the larger of \(1+\|a\|_2\) and the norms of the finitely many initial terms gives a finite bound for every term of the sequence. Hence the sequence is bounded. \(\square\)

Boundedness is necessary for convergence, but it is not sufficient. A bounded sequence may continue to move among points that are separated from one another. The next example makes that distinction explicit.

Limits in \(\mathbb{R}^n\) are unique. This is the Hausdorff-space theorem on uniqueness of limits of convergent sequences established earlier in the course: \(\mathbb{R}^n\) with its Euclidean topology is Hausdorff. We will therefore refer to “the limit” without ambiguity whenever a sequence converges.

Algebra of Limits

Vector addition and scalar multiplication interact well with convergence. These facts allow complicated sequences to be handled by combining simpler ones. The key estimate for addition is the triangle inequality: if two vectors are each close to their respective limits, their sum is close to the sum of those limits.

Proof. Fix \(\varepsilon>0\). Since \(x^{(k)}\to a\), there is an index \(K_1\) such that \(\|x^{(k)}-a\|_2<\varepsilon/2\) for \(k\geq K_1\). Since \(y^{(k)}\to b\), there is an index \(K_2\) such that \(\|y^{(k)}-b\|_2<\varepsilon/2\) for \(k\geq K_2\). For \(k\geq K=\max\{K_1,K_2\}\), the triangle inequality gives \[ \|(x^{(k)}+y^{(k)})-(a+b)\|_2 \leq\|x^{(k)}-a\|_2+\|y^{(k)}-b\|_2 <\varepsilon. \] This proves convergence of the sums.

For scalar multiplication, if \(c=0\), then \(c x^{(k)}=0=ca\) for every \(k\), so the result holds. If \(c\neq0\), choose \(K\) so that \(\|x^{(k)}-a\|_2<\varepsilon/|c|\) whenever \(k\geq K\). Homogeneity of the norm gives \[ \|c x^{(k)}-ca\|_2 =|c|\,\|x^{(k)}-a\|_2 <\varepsilon \qquad (k\geq K). \] Thus \(c x^{(k)}\to ca\). Repeated application of the two results proves the assertion for any fixed finite linear combination. \(\square\)

The Choice of Standard Norm Does Not Change Convergence

The definition used the Euclidean norm, but \(\mathbb{R}^n\) also has the standard \(1\)-norm and infinity norm. The Comparison of the Standard Norms theorem from “Norms on \(\mathbb{R}^n\)” gives, for every \(v\in\mathbb{R}^n\), \[ \|v\|_\infty\leq\|v\|_2\leq\sqrt{n}\,\|v\|_\infty, \qquad \|v\|_2\leq\|v\|_1\leq\sqrt{n}\,\|v\|_2. \] These inequalities show that requiring the distance between terms and a proposed limit to tend to zero gives the same notion for each of these norms.

Proof. Let \(a\in\mathbb{R}^n\). If \(x^{(k)}\to a\) in the Euclidean norm, then \(\|x^{(k)}-a\|_\infty\leq\|x^{(k)}-a\|_2\) and \(\|x^{(k)}-a\|_1\leq\sqrt{n}\|x^{(k)}-a\|_2\). Both right-hand sides tend to zero, so the sequence converges to \(a\) in the infinity and \(1\)-norms.

Conversely, if \(x^{(k)}\to a\) in the infinity norm, then \(\|x^{(k)}-a\|_2\leq\sqrt{n}\|x^{(k)}-a\|_\infty\), so it converges in the Euclidean norm. If it converges in the \(1\)-norm, then \(\|x^{(k)}-a\|_2\leq\|x^{(k)}-a\|_1\), so it also converges in the Euclidean norm. These implications establish equivalence. They concern convergence to each proposed \(a\); uniqueness of limits in \(\mathbb{R}^n\), noted above, ensures that different norms cannot give different limits. \(\square\)

How to Use the Definition Carefully

To prove convergence, first identify a candidate limit and then estimate the distance to it. The estimate must apply to every sufficiently large index. A typical proof chooses \(K\) in terms of the requested tolerance \(\varepsilon\), and then checks the inequality for all \(k\geq K\). Proving only that the distance is small for some indices, or that it is small for infinitely many indices, does not meet the definition.

For a nonconvergence proof, it is often effective to show that terms remain separated. In the alternating example, sufficiently late terms include two points at distance \(2\), so they cannot all lie within a ball of radius \(1/2\) around one proposed limit. More generally, if a sequence converges, every sufficiently late pair of terms must be close to one another: for any \(\varepsilon>0\), the triangle inequality shows that two terms each within \(\varepsilon/2\) of the limit are less than \(\varepsilon\) apart. Persistent separation therefore rules out convergence.

The results here establish basic tools without yet replacing vector convergence by separate conditions on its coordinates. In the next tutorial, coordinatewise convergence will be related precisely to convergence in \(\mathbb{R}^n\). For now, the norm definition provides a direct geometric test: all sufficiently late vectors must lie in every prescribed ball about the limit.