# Convergence of a sequence of averages of a convergent sequence

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• Eclair_de_XII
In summary, the conversation discusses the concept of Cesàro summation and its properties. It states that for any given positive real number epsilon, there exists an integer N such that for any positive integer n greater than or equal to N, the absolute value of the difference between the arithmetic mean of the first n terms of a sequence and its limit A is less than epsilon. It also proves that if this is not the case, then the sequence does not converge to A. This is followed by a discussion on the convergence of 2 times the arithmetic mean to 2 times A, and how this implies the convergence of the arithmetic mean itself to A. The conversation concludes by referencing the concept of Cesàro summation on Wikipedia for further information
Eclair_de_XII
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##.

## 1. What is the definition of convergence of a sequence of averages?

The convergence of a sequence of averages refers to the behavior of the average values of a sequence as the number of terms in the sequence increases. In other words, it describes whether the average values of the sequence approach a specific value as the number of terms increases.

## 2. How is the convergence of a sequence of averages related to the convergence of the original sequence?

The convergence of a sequence of averages is directly related to the convergence of the original sequence. If the original sequence converges to a specific value, then the sequence of averages will also converge to the same value. However, if the original sequence does not converge, then the sequence of averages may or may not converge.

## 3. What is the significance of the convergence of a sequence of averages?

The convergence of a sequence of averages is important because it allows us to determine the behavior of a sequence as the number of terms increases. It also helps us to identify the limit of a sequence, which is the value that the sequence approaches as the number of terms increases towards infinity.

## 4. Can a sequence of averages converge if the original sequence does not?

Yes, it is possible for a sequence of averages to converge even if the original sequence does not. This can happen if the original sequence has values that are oscillating or fluctuating around a specific value, but the average values of the sequence still approach that value as the number of terms increases.

## 5. How can we determine the convergence of a sequence of averages?

The convergence of a sequence of averages can be determined by calculating the limit of the sequence, which is the value that the sequence approaches as the number of terms increases towards infinity. If the limit exists and is a finite value, then the sequence of averages converges. If the limit does not exist or is infinite, then the sequence of averages does not converge.

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