# convergence in the mean

Let

 $b_{n}:=\frac{a_{1}+a_{2}+\ldots+a_{n}}{n}\quad(n=1,2,3,\ldots)$
 $\displaystyle a_{1},a_{2},a_{3},\ldots$ (1)

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

 $\displaystyle b_{1},b_{2},b_{3},\ldots$ (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 $\varepsilon$ be an arbitrary positive number.  We may write

 $\displaystyle|A-b_{n}|$ $\displaystyle=|A-\frac{1}{n}(a_{1}+\ldots+a_{k})-\frac{1}{n}(a_{k+1}+\ldots+a_% {n})|$ $\displaystyle=|\frac{1}{n}[(A-a_{1})+\ldots+(A-a_{k})]+\frac{1}{n}[(A-a_{k+1})% +\ldots+(A-a_{n})]|$ $\displaystyle\leqq\frac{|(A-a_{1})+\ldots+(A-a_{k})|}{n}+\frac{|A-a_{k+1}|+% \ldots+|A-a_{n}|}{n}.$

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

 $|A-a_{i}|<\frac{\varepsilon}{2}\quad\mbox{ for all }i>k.$

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

 $\frac{|(A-a_{1})+\ldots+(A-a_{k})|}{n}<\frac{\varepsilon}{2}\quad\mbox{ for }n% >l.$

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

 $|A-b_{n}|\;<\;\frac{\varepsilon}{2}+\frac{1}{n}\!(n-k)\!\!\frac{\varepsilon}{2% }\;<\;\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon,$

whence we have

 $\lim_{n\to\infty}b_{n}=A.$

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  $0,1,0,1,0,1,\ldots,$  then it is divergent but converges in the mean to the limit $\frac{1}{2}$; the corresponding sequence (2) is $0,\frac{1}{2},\frac{1}{3},\frac{2}{4},\frac{2}{5},\frac{3}{6},\frac{3}{7},% \frac{4}{8},\frac{4}{9},\ldots$

Title convergence in the mean ConvergenceInTheMean 2015-04-08 7:29:35 2015-04-08 7:29:35 pahio (2872) pahio (2872) 7 pahio (2872) Definition