Let

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

be the arithmetic mean of the numbers $a_{1},a_{2},\ldots,a_{n}$.   The sequence

$\displaystyle a_{1},a_{2},a_{3},\ldots$

(1)

is said to converge in the mean 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 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 $0,\frac{1}{2},\frac{1}{3},\frac{2}{4},\frac{2}{5},\frac{3}{6},\frac{3}{7},\frac{4}{8},\frac{4}{9},\ldots$