B Convergence of a sequence of averages of a convergent sequence

[Eclair_de_XII]

TL;DR

Consider a sequence $a_n$ and for each $n$, define:

$\alpha_n=\frac{1}{n}\sum_{i=1}^n a_i$

Prove that if $a_n$ converges to $A$, then $\alpha_n$ converges to $A$.

Let $\epsilon>0$. Then there is an integer $N>0$ with the property that for any integer $n\geq N$, $|a_n-A|<\epsilon$, where $A\in\mathbb{R}$.

If for all positive integers $n$, it is the case that $|a_n-A|<\epsilon$, then the following must hold:

\begin{eqnarray}
|\frac{1}{n}(a_1+\cdots+a_n)-A|&=&|\frac{1}{n}(a_1+\cdots+a_n)-\frac{1}{n}(A+\cdots+A)|\\
&\leq&\frac{1}{n}\sum_{i=1}^n|a_i-A|\\
&<&\frac{1}{n}\sum_{i=1}^n\epsilon\\
&=&\frac{1}{n}(n\epsilon)\\
&=&\epsilon
\end{eqnarray}

Now assume that this is not the case. Then there exists a finite set of integers $I$ s.t. if $i\in I$, then $|a_i-A|>|a_N-A|$. Denote $M=\max\{|a_i-A|:i\in I\}$.

We prove that $2\alpha_n$ converges to $2A$. Choose the integer $N'$ with the property that $N'\geq\frac{M\cdot |I|}{\epsilon}$. Then whenever $n\geq N'$:

\begin{eqnarray}
|\frac{1}{n}(a_1+\cdots+a_n)-\frac{1}{n}(A+\cdots+A)|&\leq&\frac{1}{n}\sum_{i=1}^n|a_i-A|\\
&=&\frac{1}{n}\sum_{i\in I} |a_i-A|+\frac{1}{n}\sum_{j\notin I}|a_j-A|\\
&<&\frac{|I|\cdot M}{N'}+\frac{1}{n}(n-|I|)\epsilon\\
&<&\epsilon+\epsilon
\end{eqnarray}

Since $2\alpha_n$ converges to $2A$, it must be the case that $\alpha_n$ converges to $A$.

 

[mathman]

https://en.wikipedia.org/wiki/Cesàro_summation

Discussion of the idea.