# Follow-up Question about Cesaro mean proof

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.

Proof:

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.

• where did the $x$ go in the 'triangle inequality' line? Jan 31, 2021 at 1:37
• @CSquared my apologies working on the edit! Jan 31, 2021 at 1:38
• @Trap.Lord Actually you could get this result very easily from the Toeplitz theorem. math.stackexchange.com/questions/2514778/toeplitz-theorem Jan 31, 2021 at 1:58
• @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! Jan 31, 2021 at 2:00

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.

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

• 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 Jan 31, 2021 at 1:52
• @CSquared yes of course; thats good to know Jan 31, 2021 at 1:53
• Thank you so much this was super insightful! Jan 31, 2021 at 1:58