I am trying to understand the proof behind the Cesaro mean converging. I am using https://math.stackexchange.com/a/2342856/633922 (hopefully it is also correct) as a guide because it seems very direct. I will comment on the steps I understand and where I need help.

The statement: If $(x_n)$ converges to $x$, the sum of averages $y_n=\dfrac{x_1+x_2+\cdots+x_n}{n}$ also converges to the same limit.


Since $(x_n)$ converges, given an arbitrary $\epsilon >0$, there exists an $N_1\in\mathbb{N}$ such that whenever $n\geq N_1$ we have $|x_n-x|<\epsilon$. (Definition of convergent sequence)

Now, $$\begin{align*} \left|\frac{x_1+x_2+\cdots+x_n}{n}-x\right|=&\left|\frac{(x_1-x)+\cdots+(x_{N_1-1}-x)}{n}+\frac{(x_{N_1}-x)+\cdots+(x_{n}-x)}{n}\right|\\ \leq& \left|\frac{(x_1-x)+\cdots+(x_{N_1-1}-x)}{n}\right|+\left|\frac{(x_{N_1}-x)+\cdots+(x_{n}-x)}{n}\right|\text{ (Triangle inequality)} \end{align*} $$

Now we want to make a statement about the first $N_1-1$ terms, $\color{red}{why?}$ That is:

By the Archimedean principle we can find an $N_2$ such that whenever $n\geq N_2$ we have that

$$\left|\frac{x_1+x_2+\cdots+x_{N-1}}{n}\right|<\epsilon $$ (Thought: is it because $x_1,\dots,x_{N_1-1}$ is finite?)

Now we can choose an $N_3=\max\{N_1,N2\}$ such that for all $n\geq N_3$ we have (My thought: Is this because choosing the max of both will always guarantee the final inequality to always work?) $$ \left|\frac{x_1+x_2+\cdots+x_n}{n}-x\right|\leq \underbrace{\epsilon}_{N_1-1}+\underbrace{\color{red}{\frac{n-N_1}{n}}}_{\text{why and how?}}\epsilon< 2\epsilon $$

And this finishes the proof. I always assumed the ending statement has to be (something)$<\epsilon$ or is this saying that each sum of the right side of the triangle inequality is less than $\epsilon/2$. I would really appreciate the help on the areas I am thoroughly confused about.

  • $\begingroup$ where did the $x$ go in the 'triangle inequality' line? $\endgroup$
    – C Squared
    Jan 31, 2021 at 1:37
  • 1
    $\begingroup$ @CSquared my apologies working on the edit! $\endgroup$ Jan 31, 2021 at 1:38
  • $\begingroup$ @Trap.Lord Actually you could get this result very easily from the Toeplitz theorem. math.stackexchange.com/questions/2514778/toeplitz-theorem $\endgroup$
    – Valent
    Jan 31, 2021 at 1:58
  • $\begingroup$ @Valerin I appreciate the extra wisdom, but I am currently in a course that is designed for people heading into real analysis, so I am not familar with this theorem but I will keep it in mind for the future! $\endgroup$ Jan 31, 2021 at 2:00

2 Answers 2

  1. The first part is controlled because $N_1$ is fixed and the denominator $n$ can be arbitrarily large.
  2. For the second part you are using the estimate $|x_n-x|<\epsilon$ for large $n$. Note that there are totally $n-N_1$ terms in the numerator, the absolute value of each is less than $\epsilon$.

To summarize, you are first given $\epsilon>0$. Then you have $N_1$ such that $n\ge N_1$ implies $|x_n-x|<\epsilon$, in particular $$ |x_{N_1+1}-x|<\epsilon,\quad |x_{N_1+2}-x|<\epsilon,\cdots,|x_{n}-x|<\epsilon\tag{1} $$ which implies by the triangle inequality that $$ \left|\frac{(x_{N_1+1}-x)+\cdots+(x_{n}-x)}{n}\right|<\frac{(n-N_1)\epsilon}{n}\tag{1'} $$

Then, for this (fixed) $N_1$, since $$ \lim_{n\to\infty}\frac{|(x_1-x)+\cdots+(x_{N_1}-x)|}{n}=0 $$ there exists $N_2$ such that $n\ge N_2$ implies $$ \frac{|(x_1-x)+\cdots+(x_{N_1}-x)|}{n}<\epsilon\tag{2} $$

So if you pick $N_3=\max(N_1,N_2)$, then $n\ge N_3$ implies both (1') and (2).

If you can show that for every $\epsilon>0$, there exists $N>0$ such that $n\ge N$ implies $|a_n|<2\epsilon$, it follows that $$ \lim_{n\to\infty}a_n=0 $$ You don't have to have exactly $\epsilon$ in the estimate.

  • $\begingroup$ Oh wow this is amazing! (1) really made me see why this works!!! For the last line $n\not=N_3$ is implying $n>N_3$? $\endgroup$ Jan 31, 2021 at 1:58
  • $\begingroup$ @Trap.Lord: that was a typo. Fixed. Thanks. :-) $\endgroup$
    – user9464
    Jan 31, 2021 at 1:59
  • $\begingroup$ Awesome! Thank you you once again! $\endgroup$ Jan 31, 2021 at 2:01
  • $\begingroup$ @Trap.Lord: you are welcome! $\endgroup$
    – user9464
    Jan 31, 2021 at 2:01

first red text: You want to make a seperate statement about the first $N_1-1$ terms since you don‘t know how the sequence behaves in those terms. You know that $|x_n-x|<\varepsilon$ for all $n\geq N_1$ but you don‘t know what‘s going on earlier in the sequence.

second red text: this follows again from the triangle inequality: $$ \left| \frac{x_{N_1}+x_{N_1+1}+\ldots+x_n}{n} - x \right| = \left| \frac{x_{N_1}-x+x_{N_1+1}-x+\ldots+x_n-x}{n} \right| \leq \frac{|x_{N_1}-x|+|x_{N_1+1}-x|+\ldots+|x_n-x|}{n} $$

that are $N_1-n$ terms of value $<\varepsilon$ (that’s because of how we chose $N_1$) in the numerator giving you the term $$ \frac{n-N_1}{n}\varepsilon $$

edit: regarding the term $2\varepsilon$: it is only important that we can make this expression arbitrarily small. Although it is „convention“ to use $\varepsilon$ it is also fine to use $2\varepsilon$ since we can make this expression as small as we want if we just make $\varepsilon$ small enough.

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    $\begingroup$ this can be modified to have $<\varepsilon$ in the end if we make $|x-x_n|<\varepsilon/2$ for $n\geq N_1$ and $$\bigg|\frac{x_1+x_2+\dots+x_{N-1}}{n}\bigg|<\varepsilon/2 $$ for $n\geq N_2$, just if OP was wondering $\endgroup$
    – C Squared
    Jan 31, 2021 at 1:52
  • $\begingroup$ @CSquared yes of course; thats good to know $\endgroup$ Jan 31, 2021 at 1:53
  • $\begingroup$ Thank you so much this was super insightful! $\endgroup$ Jan 31, 2021 at 1:58

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