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Show that if $\lim x_n =x$ with $x_n \in \mathbb{R}$ and $x \in \mathbb{R}$, then the sequence given by the averages

$y_n = \frac{x_1 +x_2 + \dots + x_n}{n}$ also converges to $x$.

Solution: Denote the limit of $(x_n)$ as $a$. Let $\varepsilon >0$ Then there exists a positive integer $N_1 > 0$ such that for all $n$,

$$|x_n-a| \leq \frac \varepsilon 2.$$

It follows that for any $n \geq N_1$, we have

\begin{align} & |y_n -a| = \left| \frac{1}{n} \sum_{i=1}^{N_1 =1}(x_i-a) + \frac{1}{n} \sum_{i=N_1 +1 }^n(x_i-a)\right| \\[8pt] \leq {} & \frac{1}{n} \sum_{i=1}^{N_1 =1}|(x_i-a)| +\frac{1}{n} \sum_{i=N_1 +1 }^n|(x_i-a)| \\[8pt] \leq {} & \frac{1}{n} \sum_{i=1}^{N_1 =1}|(x_i-a)| + \frac{(n-N_1)}{n} \cdot \frac{\varepsilon}{2} \\[8pt] \leq {} & \frac{1}{n} \sum_{i=1}^{N_1 =1}|(x_i-a)|+ \frac{\varepsilon}{2} \end{align}

Let $M=\frac{1}{n} \sum_{i=1}^{N_1 =1}|(x_i-a)|$. This is where i am lost because I know that there exists a positive integer $N_2 >0$ such that for all $n \geq N_2$ but i dont know why. This would mean

$$\frac{1}{n} \sum_{i=1}^{N_1 =1}|(x_i-a)| \leq \frac{\varepsilon}{2}.$$

Now by combining this with the result from above, if $n \geq \max\{N_1,N_2\}$, then

$$| y_n - a| \leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$

Michael Hardy
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SelfStudy
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1 Answers1

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You have chosen $N_1$ already, so $\sum_{i=1}^{N_1} |x_i-a|$ is a constant $C$. So, we can choose $N_2 \ge N_1$ large enough so that $\frac{1}{n} \sum_{i=1}^{N_1} |x_i-a| = \frac{C}{n}< \epsilon/2$ for any $n \ge N_2$.

angryavian
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  • thanks for your comment, but could you please explain why it is so that 'we can choose...for any n greater than or equal to $N_2$? – Pianoman1234 May 20 '18 at 16:20