- #1

Eclair_de_XII

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- TL;DR Summary
- 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##.

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##.