Sequences can appear to settle down long before we understand what they are approaching. The decimals $3$, $3.1$, $3.14$, $3.141$ suggest a limiting number, while $(-1)^n$ never chooses a side. Convergence is the language that separates genuine stabilization from repeated pattern.

The main problem is not only whether a limit exists. The deeper question is what survives after passing to that limit. Continuity, inequalities, integrals, derivatives, and minimizers are all fragile under weak forms of convergence. A useful theory must say which mode of convergence protects which operation.

[example: A Pointwise Limit That Breaks Continuity] Here $C([0,1];\mathbb{R})$ denotes the set of continuous functions from $[0,1]$ to $\mathbb{R}$. For each $n \in \mathbb{N}$, define $f_n\in C([0,1];\mathbb{R})$ by \begin{align*} f_n(x)=x^n,\qquad 0\le x\le 1. \end{align*} We compute the pointwise limit. If $x=0$, then $f_n(0)=0^n=0$ for every $n$, so $f_n(0)\to 0$. If $0<x<1$, write $x=1/(1+t)$ with $t=(1-x)/x>0$. For every $n\in\mathbb{N}$, Bernoulli's inequality gives $(1+t)^n\ge 1+nt$, hence \begin{align*} 0\le x^n=\frac{1}{(1+t)^n}\le \frac{1}{1+nt}. \end{align*} Given $\varepsilon>0$, choose $N\in\mathbb{N}$ with $N>(1/\varepsilon-1)/t$ when $\varepsilon<1$, and take $N=1$ when $\varepsilon\ge 1$. Then for every $n\ge N$, \begin{align*} 0\le x^n\le \frac{1}{1+nt}<\varepsilon. \end{align*} Thus $x^n\to 0$ for every $0\le x<1$. At the endpoint, \begin{align*} f_n(1)=1^n=1 \end{align*} for every $n$, so $f_n(1)\to 1$. Therefore the pointwise limit $f:[0,1]\to\mathbb{R}$ is \begin{align*} f(x)=0\text{ for }0\le x<1,\qquad f(1)=1. \end{align*} This limit is not continuous at $1$: for $\varepsilon=1/2$, every $\delta>0$ admits a point $y\in[0,1)$ with $|y-1|<\delta$, for instance $y=1-\min(\delta/2,1/2)$, and then \begin{align*} |f(y)-f(1)|=|0-1|=1>\frac12. \end{align*} So each $f_n$ is continuous, but the pointwise limit is discontinuous; pointwise convergence controls each fixed input separately and does not preserve continuity. [/example]

The failure is concentrated near $1$. For each fixed $x<1$ the sequence eventually becomes small, but there is no single stage after which all points near $1$ are controlled. This distinction between point-by-point control and uniform control will guide the whole chapter.

## Definition

To speak about convergence, we first need a structure that measures closeness. In $\mathbb{R}$ this is absolute value; in a general [metric space](/page/Metric%20Space) it is a distance function. The definition asks whether every tolerance around a candidate limit eventually traps the sequence.

[definition: Convergence in a Metric Space] Let $(X,d)$ be a metric space, let $(x_n)_{n=1}^\infty$ be a sequence in $X$, and let $x\in X$. The sequence $(x_n)$ converges to $x$ if for every $\varepsilon>0$ there exists $N\in\mathbb{N}$ such that for every $n\ge N$, \begin{align*} d(x_n,x)<\varepsilon. \end{align*} Write $x_n\to x$ as $n\to\infty$. [/definition]

This definition depends on a proposed limit. In many constructions, however, an approximation scheme gives terms that are close to each other without revealing the limiting object in advance. We therefore need an internal convergence test that mentions only the tail of the sequence.

## Cauchy Sequences and Completeness

When the candidate limit is unknown, the previous definition is hard to use directly. A numerical scheme, an iteration, or a sequence of approximations may only tell us that late terms are close to one another. The Cauchy condition isolates this internal stabilization: the tail must eventually fit inside any prescribed tolerance, even before we know where it should land.

[definition: Cauchy Sequence] Let $(X,d)$ be a metric space. A sequence $(x_n)_{n=1}^\infty$ in $X$ is a [Cauchy sequence](/page/Cauchy%20Sequence) if for every $\varepsilon>0$ there exists $N\in\mathbb{N}$ such that for all $m,n\ge N$, \begin{align*} d(x_m,x_n)<\varepsilon. \end{align*} [/definition]

A Cauchy sequence may still fail to converge if the ambient space has holes. The rational approximations to $\sqrt{2}$ want to converge, but their limit is absent from $\mathbb{Q}$. This motivates making completeness a property of the space itself.

[definition: Complete Metric Space] A metric space $(X,d)$ is complete if every Cauchy sequence in $X$ converges to an element of $X$. [/definition]

Completeness matters because existence proofs often build Cauchy sequences rather than closed formulas. Before completeness can be used, we need the basic fact that genuine convergence always gives the Cauchy property.

[quotetheorem:7844]

The theorem gives a necessary test for convergence. It also explains why completeness is the missing converse condition: in a complete space, the internal Cauchy test is enough.

[example: A Cauchy Sequence Missing Its Limit in $\mathbb{Q}$] Let \begin{align*}q_n=\frac{\lfloor 10^n\sqrt{2}\rfloor}{10^n}\in\mathbb{Q}.\end{align*} By the defining property of the floor function, \begin{align*}\lfloor 10^n\sqrt{2}\rfloor\le 10^n\sqrt{2}<\lfloor 10^n\sqrt{2}\rfloor+1.\end{align*} Dividing by $10^n$ gives \begin{align*}q_n\le \sqrt{2}<q_n+10^{-n},\end{align*} so \begin{align*}0\le \sqrt{2}-q_n<10^{-n}.\end{align*} We show that $(q_n)$ is Cauchy in $\mathbb{Q}$ with the metric $d(x,y)=|x-y|$. Let $\varepsilon>0$, and choose $N\in\mathbb{N}$ such that $2\cdot 10^{-N}<\varepsilon$. If $m,n\ge N$, then \begin{align*}|q_m-q_n|=|(\sqrt{2}-q_n)-(\sqrt{2}-q_m)|.\end{align*} By the triangle inequality and the error bound above, \begin{align*}|q_m-q_n|\le |\sqrt{2}-q_n|+|\sqrt{2}-q_m|<10^{-n}+10^{-m}\le 2\cdot 10^{-N}<\varepsilon.\end{align*} Thus $(q_n)$ is Cauchy in $\mathbb{Q}$. It remains to see why the sequence has no limit in $\mathbb{Q}$. Suppose, toward a contradiction, that $q_n\to q$ in $\mathbb{Q}$ for some $q\in\mathbb{Q}$. The same absolute-value metric is being used inside $\mathbb{R}$, so also $q_n\to q$ as a real sequence. But the bound $0\le \sqrt{2}-q_n<10^{-n}$ shows $q_n\to\sqrt{2}$ in $\mathbb{R}$: given $\varepsilon>0$, choose $N$ with $10^{-N}<\varepsilon$, and then $n\ge N$ implies $|q_n-\sqrt{2}|<\varepsilon$. A real sequence has at most one limit: if $q_n\to q$ and $q_n\to \sqrt{2}$ with $q\ne\sqrt{2}$, take $\varepsilon=|q-\sqrt{2}|/3$ and choose $n$ large enough that $|q_n-q|<\varepsilon$ and $|q_n-\sqrt{2}|<\varepsilon$; then \begin{align*}|q-\sqrt{2}|\le |q-q_n|+|q_n-\sqrt{2}|<2\varepsilon=\frac{2}{3}|q-\sqrt{2}|,\end{align*} which is impossible. Hence $q=\sqrt{2}$. Finally, $\sqrt{2}$ is not rational. If $\sqrt{2}=a/b$ with integers $a,b$ having no common factor and $b\ne 0$, then $a^2=2b^2$, so $a^2$ is even and therefore $a$ is even. Write $a=2c$. Then \begin{align*}4c^2=2b^2.\end{align*} Dividing by $2$ gives \begin{align*}2c^2=b^2,\end{align*} so $b$ is even, contradicting that $a$ and $b$ have no common factor. Therefore $(q_n)$ is Cauchy in $\mathbb{Q}$ but does not converge in $\mathbb{Q}$; the missing limit is the real number $\sqrt{2}$. [/example]

## Numerical Limits and Algebraic Stability

### Limit Laws

Numerical convergence is the first setting where limits become calculational. We want to replace complicated sequences by their limits inside algebraic expressions, but this is legitimate only when the operations are continuous with respect to the chosen notion of closeness.