convergence in the mean

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 (http://planetmath.org/ConvergenceInTheMean) iff thesequence

\\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 nottrue. 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 $0,\\frac{1}{2},\\frac{1}{3},\\frac{2}{4},\\frac{2}{5},\\frac{3}{6},\\frac{3}{7},%\\frac{4}{8},\\frac{4}{9},\\ldots$