What is Convergent sequence? — Mathematics

Definition 20.10 High School Mathematics · Chapter 20 — Sequences

A sequence $(u_n)$ converges to the real number $\ell$ if every open interval containing $\ell$ contains all the terms $u_n$ from some index on. We then write $\lim\limits_{n\to+\infty} u_n = \ell$ .

Equivalently: for every $\varepsilon > 0$ , there exists $N \in \N$ such that for all $n \geq N$ , $\abs{u_n - \ell} \leq \varepsilon$ .

Convergence of u_n = 2 + (-1)n/n to = 2: given > 0, all terms from the index N on lie in the band [ - , + ]. — Convergence of $u_n = 2 + \frac{(-1)^n}{n}$ to $\ell = 2$ : given $\varepsilon > 0$ , all terms from the index $N$ on lie in the band $\intcc{\ell-\varepsilon}{\ell+\varepsilon}$ .

Examples

Example 20.22

Let $u_0 = 2$ and $u_{n+1} = \frac{1}{2}\left(u_n + \frac{2}{u_n}\right)$ . One checks by induction that $u_n \geq \sqrt{2}$ for all $n$ (the inequality $\frac{1}{2}(x + 2/x) \geq \sqrt{2}$ for $x>0$ is equivalent to $(x - \sqrt2)^2 \geq 0$ ), then that $(u_n)$ is decreasing, since

u_{n+1}-u_n=\frac{2-u_n^2}{2u_n}\leq 0 .

Decreasing and bounded below, $(u_n)$ converges to some $\ell \geq \sqrt{2}$ , which must satisfy $\ell = \frac{1}{2}(\ell + 2/\ell)$ , i.e. $\ell^2 = 2$ . Hence $u_n \to \sqrt{2}$ . This is Heron’s algorithm, already used by the Babylonians; its convergence is extremely fast ( $u_3$ already gives $\sqrt 2$ to eight decimal places).

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Examples

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