Difference between revisions of "Convergence of a sequence"
(Created page with "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 {{...") |
m |
||
Line 1: | Line 1: | ||
+ | {{Refactor notice}} | ||
+ | '''This page is to be phased out and the content moved to either more appropriate places or to [[Limit (sequence)]]''' | ||
+ | |||
Like with [[Continuous map|continuity]] there are three forms for convergence of a [[Sequence]] | Like with [[Continuous map|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 {{M|a}} or <math>\lim_{n\rightarrow\infty}(a_n)=a</math> if and only if the following definition holds: | Given a sequence <math>(a_n)^n_{n=1}</math> we may say that it converges to {{M|a}} or <math>\lim_{n\rightarrow\infty}(a_n)=a</math> if and only if the following definition holds: | ||
+ | |||
+ | __TOC__ | ||
==First form== | ==First form== | ||
Introductory form | Introductory form |
Revision as of 15:29, 24 November 2015
This page is to be phased out and the content moved to either more appropriate places or to Limit (sequence)
Like with continuity there are three forms for convergence of a Sequence
Given a sequence (an)nn=1 we may say that it converges to a or lim if and only if the following definition holds:
Contents
[hide]First form
Introductory form \forall\epsilon>0\exists N\in\mathbb{N}:n> N\implies|a_n-a|<\epsilon
Second form
Metric space form \forall\epsilon>0\exists N\in\mathbb{N}:n> N\implies d(a_n-a)<\epsilon
Third form
Topological form \forall N_a\exists N\in\mathbb{N}: n> N\implies a_n\in N_a where N_a denotes a neighbourhood of a
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 d_{\|\cdot\|}(x,y)=\|x-y\|, thus we can also have a slight variation of the metric form:
\forall\epsilon>0\exists N\in\mathbb{N}:n> N\implies \|a_n-a\|<\epsilon
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
f_n(t)=t^n\rightarrow 0 in \|\cdot\|_{L^1}
Using the \|\cdot\|_{L^1} norm stated here for convenience: \|f\|_{L^p}=\left(\int^1_0|f(x)|^pdx\right)^\frac{1}{p} so \|f\|_{L^1}=\int^1_0|f(x)|dx
We see that \|f_n\|_{L^1}=\int^1_0x^ndx=\left[\frac{1}{n+1}x^{n+1}\right]^1_0=\frac{1}{n+1}
This clearly \rightarrow 0 - this is 0:[0,1]\rightarrow\mathbb{R} which of course has norm 0, we think of this from the sequence (\|f_n-0\|_{L^1})^\infty_{n=1}\rightarrow 0\iff f_n\rightarrow 0