Convergence of a sequence

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Like with continuity there are three forms for convergence of a Sequence

Given a sequence [math](a_n)^n_{n=1}[/math] we may say that it converges to [ilmath]a[/ilmath] or [math]\lim_{n\rightarrow\infty}(a_n)=a[/math] if and only if the following definition holds:

First form

Introductory form [math]\forall\epsilon>0\exists N\in\mathbb{N}:n> N\implies|a_n-a|<\epsilon[/math]

Second form

Metric space form [math]\forall\epsilon>0\exists N\in\mathbb{N}:n> N\implies d(a_n-a)<\epsilon[/math]

Third form

Topological form [math]\forall N_a\exists N\in\mathbb{N}: n> N\implies a_n\in N_a[/math] where [math]N_a[/math] denotes a neighbourhood of [math]a[/math]

Cauchy Criterion

Convergence can be shown without knowing what exactly the sequence converges to, see the Cauchy criterion for convergence page

Note on norms

Recall from norm that we can simply define [math]d_{\|\cdot\|}(x,y)=\|x-y\|[/math], thus we can also have a slight variation of the metric form:

[math]\forall\epsilon>0\exists N\in\mathbb{N}:n> N\implies \|a_n-a\|<\epsilon[/math]

Is is worth noting because in Functional Analysis norms are considered and if we deal with a metric space we are inside a branch of topology

Interesting examples

[math]f_n(t)=t^n\rightarrow 0[/math] in [math]\|\cdot\|_{L^1}[/math]

Using the [math]\|\cdot\|_{L^1}[/math] norm stated here for convenience: [math]\|f\|_{L^p}=\left(\int^1_0|f(x)|^pdx\right)^\frac{1}{p}[/math] so [math]\|f\|_{L^1}=\int^1_0|f(x)|dx[/math]

We see that [math]\|f_n\|_{L^1}=\int^1_0x^ndx=\left[\frac{1}{n+1}x^{n+1}\right]^1_0=\frac{1}{n+1}[/math]

This clearly [math]\rightarrow 0[/math] - this is [math]0:[0,1]\rightarrow\mathbb{R}[/math] which of course has norm [ilmath]0[/ilmath], we think of this from the sequence [math](\|f_n-0\|_{L^1})^\infty_{n=1}\rightarrow 0\iff f_n\rightarrow 0[/math]