# convergence in the mean

## Primary tabs

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\begin{document}
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
\begin{align}
a_1, a_2, a_3, \ldots
\end{align}
is said to \PMlinkname{{\it converge in the mean}}{ConvergenceInTheMean} iff the
sequence
\begin{align}
b_1,b_2,b_3,\ldots
\end{align}
converges.\\
On has the

\textbf{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.

{\it Proof.}\, Let $\varepsilon$ be an arbitrary positive number.\, We may write
\begin{align*}
|A-b_n|  &= |A-\frac{1}{n}(a_1+\ldots+a_k)-\frac{1}{n}(a_{k+1}+\ldots+a_n)|\\
&= |\frac{1}{n}[(A-a_1)+\ldots+(A-a_k)]+\frac{1}{n}[(A-a_{k+1})+\ldots+(A-a_n)]|\\
&\leqq \frac{|(A-a_1)+\ldots+(A-a_k)|}{n}+\frac{|A-a_{k+1}|+\ldots+|A-a_n|}{n}.
\end{align*}
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\,
$$\lim_{n\to\infty}b_n = A.$$\\

\textbf{Note.}\, The \PMlinkname{converse}{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
$0,\frac{1}{2},\frac{1}{3},\frac{2}{4},\frac{2}{5}, \frac{3}{6},\frac{3}{7},\frac{4}{8},\frac{4}{9},\ldots$\\

nd{document}