convergence in the mean

Let

$$b_n:=\frac{a_1+a_2+\mathrm{\dots }+a_n}{n}\mathit{\hspace{1em}}(n=1,2,3,\mathrm{\dots })$$

be the arithmetic mean of the numbers $a_1,a_2,\mathrm{\dots },a_n$.   The sequence

$a_1,a_2,a_3,\mathrm{\dots }$ (1)

is said to converge in the mean (http://planetmath.org/ConvergenceInTheMean) iff the sequence

$b_1,b_2,b_3,\mathrm{\dots }$ (2)

converges.
On has the

Theorem.  If the sequence (1) is convergent having the limit $A$, then also the sequence (2) converges to the limit $A$.  Thus, a convergent sequence is always convergent in the mean.

Proof.  Let $\epsilon$ be an arbitrary positive number.  We may write

$|A-b_n|$ $=|A-{\displaystyle \frac{1}{n}}(a_1+\mathrm{\dots }+a_k)-{\displaystyle \frac{1}{n}}(a_{k+1}+\mathrm{\dots }+a_n)|$

$=|{\displaystyle \frac{1}{n}}[(A-a_1)+\mathrm{\dots }+(A-a_k)]+{\displaystyle \frac{1}{n}}[(A-a_{k+1})+\mathrm{\dots }+(A-a_n)]|$

$\leqq {\displaystyle \frac{|(A-a_1)+\mathrm{\dots }+(A-a_k)|}{n}}+{\displaystyle \frac{|A-a_{k+1}|+\mathrm{\dots }+|A-a_n|}{n}}.$

The supposition implies that there is a positive integer $k$ such that

Let’s fix the integer $k$.  Choose the number $l$ so great that

Let now  $n>\max \{k,l\}$.  The three above inequalities yield

whence we have

$$\underset{n\to \mathrm{\infty }}{lim}{b}_n=A.$$

Note.  The converse (http://planetmath.org/Converse) of the theorem is not true.  For example, if

$$a_n:=\frac{1+{(-1)}^n}{2}$$

i.e. if the sequence (1) has the form  $0,1,0,1,0,1,\mathrm{\dots },$  then it is divergent but converges in the mean to the limit $\frac{1}{2}$; the corresponding sequence (2) is $0,\frac{1}{2},\frac{1}{3},\frac{2}{4},\frac{2}{5},\frac{3}{6},\frac{3}{7},\frac{4}{8},\frac{4}{9},\mathrm{\dots }$

Title	convergence in the mean
Canonical name	ConvergenceInTheMean
Date of creation	2015-04-08 7:29:35
Last modified on	2015-04-08 7:29:35
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	7
Author	pahio (2872)
Entry type	Definition