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#2: Post edited
by
TextKit
·
2021-03-11T00:21:07Z (almost 4 years ago)
- >**Exercise 2.3.11 (Cesaro Means).** (a) Show if $\{x_n\}$ is a convergent sequence, then the sequences given by the averages $\{\dfrac{x_1 + x_2 + ... + x_n}{n}\}$ converges to the same limit.
I rewrote and colored the official solution.- >Let $\epsilon>0$ be arbitrary. Then we need to find an $N \in \mathbb{N} \qquad \ni n \geq N \implies\ |\frac{x_1 + x_2 + ... + x_n}{n} - L|< \epsilon \tag{1}$.
- >Question posits $(x_{n}) \to L$. So $\exists \; M \in \mathbb{N} \ni n \ge M \implies |x_{n}-L|< M \quad (2)$.
- >$\text{Also } \exists \; C \ni n \ge C \implies |x_{n}-L|< \epsilon/2. \quad \tag{3}$
- **My question 1.** Where does (3) come from? How to prognosticate $\epsilon/2$? Normally you start with $\epsilon$.
- 2. Doesn't the same argument prove all the terms can be bounded? Why simply ['the early terms in the averages can be bounded'](https://math.stackexchange.com/a/700667) ? PWhy "Because the original sequence is convergent, we suspect that we can bound ... we will be breaking the limit in two at the end" ?
- I still don't understand how "we suspect" these bounds?
- >Now for all $n \ge C$, we can write
- ![enter image description here][1]
3. How to presage rewriting $\color{red}{L = nL/n}?$- 4. How to presage splitting the sum between $x_{C - 1}$ and $x_C$?
Now apply the Triangle Inequality to each of the $n$ $(x_i - L)$ terms.- >$\begin{align}
- \le & \frac{1}{n}( & \color{green}{\left| x_{1}-L\right| +\ldots +\left| x_{c-1}-L\right|} & + \color{brown}{\left| x_{C}-L\right| +\ldots +\left| x_{n}-L\right|} & )\\
- & & \color{green}{\text{Each of these $(C - 1)$ terms < M by (2)} } & \color{brown}{ \text{ Each of these $(n - C)$ terms < $\epsilon/2$ by (3)}} & \\
- \le & \frac{1}{n}( & \color{green}{(C - 1)M} & + \color{brown}{\frac{e}{2}(n - C)} & ). \tag{4}\\
- \end{align}$
- 5. Why's the last inequality (4) above $\le$? Why not $<$ like (2) and (3)?
- >Because C and M are fixed constants at this point, we may choose $N_2$ so that
- $\color{green}{(C - 1)M}\frac{1}{n} < e/2 \tag{5}$ for all $n \ge N_2$. Finally, let $N$ [in $(1)$] $= \max\{C, N_2\}$ be the desired $N$.
- 4. Why are we authorized to choose $N_2$ so that $\color{green}{(C - 1)M}\frac{1}{n} < e/2$?
>$C \in \mathbb{N} \quad \therefore n - C < n \iff \color{magenta}{\frac{n - C}{n}} < 1 \tag{6}$- >$\begin{align}
- \text{Equation (4) is} & & \color{green}{(C - 1)M}\frac{1}{n} & + \frac{e}{2}\color{magenta}{\frac{n - C}{n}}. \\
- \text{Then by (5) and (6),} & & < e/2 & + e/2\color{magenta}{(1)}/ QED. \\
- \end{align}$
- [Stephen Abbott](http://www.middlebury.edu/academics/math/faculty/node/26391). [*Understanding Analysis* (2016 2 edn)](https://www.amazon.com/Understanding-Analysis-Stephen-Abbott/dp/1493927116). p. 55.
- [1]: http://i.stack.imgur.com/VGDbn.png
- >**Exercise 2.3.11 (Cesaro Means).** (a) Show if $\{x_n\}$ is a convergent sequence, then the sequences given by the averages $\{\dfrac{x_1 + x_2 + ... + x_n}{n}\}$ converges to the same limit.
- I rewrote and colored [the official solution](https://math.codidact.com/uploads/YRP7Xiqfj7b9mEDWtD74bRYX).
- >Let $\epsilon>0$ be arbitrary. Then we need to find an $N \in \mathbb{N} \qquad \ni n \geq N \implies\ |\frac{x_1 + x_2 + ... + x_n}{n} - L|< \epsilon \tag{1}$.
- >Question posits $(x_{n}) \to L$. So $\exists \; M \in \mathbb{N} \ni n \ge M \implies |x_{n}-L|< M \quad (2)$.
- >$\text{Also } \exists \; C \ni n \ge C \implies |x_{n}-L|< \epsilon/2. \quad \tag{3}$
- **My question 1.** Where does (3) come from? How to prognosticate $\epsilon/2$? Normally you start with $\epsilon$.
- 2. Doesn't the same argument prove all the terms can be bounded? Why simply ['the early terms in the averages can be bounded'](https://math.stackexchange.com/a/700667) ? PWhy "Because the original sequence is convergent, we suspect that we can bound ... we will be breaking the limit in two at the end" ?
- I still don't understand how "we suspect" these bounds?
- >Now for all $n \ge C$, we can write
- ![enter image description here][1]
- 3. How to presage rewriting $\color{red}{L = nL/n}?$ and subtracting $\color{red}{L}?$ from the $x_i$?
- 4. How to presage splitting the sum between $x_{C - 1}$ and $x_C$?
- Now apply the Triangle Inequality to each of the $n$ $(x_i - L)$ terms. **Can you please correct the MathJax below? It works on S.E. The second line of inequality is supposed to align with the first. Please remove this once corrected.**
- >$\begin{align}
- \le & \frac{1}{n}( & \color{green}{\left| x_{1}-L\right| +\ldots +\left| x_{c-1}-L\right|} & + \color{brown}{\left| x_{C}-L\right| +\ldots +\left| x_{n}-L\right|} & )\\
- & & \color{green}{\text{Each of these $(C - 1)$ terms < M by (2)} } & \color{brown}{ \text{ Each of these $(n - C)$ terms < $\epsilon/2$ by (3)}} & \\
- \le & \frac{1}{n}( & \color{green}{(C - 1)M} & + \color{brown}{\frac{e}{2}(n - C)} & ). \tag{4}\\
- \end{align}$
- 5. Why's the last inequality (4) above $\le$? Why not $<$ like (2) and (3)?
- >Because C and M are fixed constants at this point, we may choose $N_2$ so that
- $\color{green}{(C - 1)M}\frac{1}{n} < e/2 \tag{5}$ for all $n \ge N_2$. Finally, let $N$ [in $(1)$] $= \max\{C, N_2\}$ be the desired $N$.
- 4. Why are we authorized to choose $N_2$ so that $\color{green}{(C - 1)M}\frac{1}{n} < e/2$?
- **Can you please correct the MathJax below? It works on S.E. The second line of inequality is supposed to align with the first. Please remove this once corrected.**
- >∵ $C \in \mathbb{N} \quad \therefore n - C < n \iff \color{magenta}{\frac{n - C}{n}} < 1 \tag{6}$
- >$\begin{align}
- \text{Equation (4) is} & & \color{green}{(C - 1)M}\frac{1}{n} & + \frac{e}{2}\color{magenta}{\frac{n - C}{n}}. \\
- \text{Then by (5) and (6),} & & < e/2 & + e/2\color{magenta}{(1)}/ QED. \\
- \end{align}$
- [Stephen Abbott](http://www.middlebury.edu/academics/math/faculty/node/26391). [*Understanding Analysis* (2016 2 edn)](https://www.amazon.com/Understanding-Analysis-Stephen-Abbott/dp/1493927116). p. 55.
- [1]: http://i.stack.imgur.com/VGDbn.png
#1: Initial revision
by
TextKit
·
2021-03-09T06:12:46Z (almost 4 years ago)
De-mystifying tricks – If $\{x_n\}$ converges, then Cesaro Mean converges.
>**Exercise 2.3.11 (Cesaro Means).** (a) Show if $\{x_n\}$ is a convergent sequence, then the sequences given by the averages $\{\dfrac{x_1 + x_2 + ... + x_n}{n}\}$ converges to the same limit.
I rewrote and colored the official solution.
>Let $\epsilon>0$ be arbitrary. Then we need to find an $N \in \mathbb{N} \qquad \ni n \geq N \implies\ |\frac{x_1 + x_2 + ... + x_n}{n} - L|< \epsilon \tag{1}$.
>Question posits $(x_{n}) \to L$. So $\exists \; M \in \mathbb{N} \ni n \ge M \implies |x_{n}-L|< M \quad (2)$.
>$\text{Also } \exists \; C \ni n \ge C \implies |x_{n}-L|< \epsilon/2. \quad \tag{3}$
**My question 1.** Where does (3) come from? How to prognosticate $\epsilon/2$? Normally you start with $\epsilon$.
2. Doesn't the same argument prove all the terms can be bounded? Why simply ['the early terms in the averages can be bounded'](https://math.stackexchange.com/a/700667) ? PWhy "Because the original sequence is convergent, we suspect that we can bound ... we will be breaking the limit in two at the end" ?
I still don't understand how "we suspect" these bounds?
>Now for all $n \ge C$, we can write
![enter image description here][1]
3. How to presage rewriting $\color{red}{L = nL/n}?$
4. How to presage splitting the sum between $x_{C - 1}$ and $x_C$?
Now apply the Triangle Inequality to each of the $n$ $(x_i - L)$ terms.
>$\begin{align}
\le & \frac{1}{n}( & \color{green}{\left| x_{1}-L\right| +\ldots +\left| x_{c-1}-L\right|} & + \color{brown}{\left| x_{C}-L\right| +\ldots +\left| x_{n}-L\right|} & )\\
& & \color{green}{\text{Each of these $(C - 1)$ terms < M by (2)} } & \color{brown}{ \text{ Each of these $(n - C)$ terms < $\epsilon/2$ by (3)}} & \\
\le & \frac{1}{n}( & \color{green}{(C - 1)M} & + \color{brown}{\frac{e}{2}(n - C)} & ). \tag{4}\\
\end{align}$
5. Why's the last inequality (4) above $\le$? Why not $<$ like (2) and (3)?
>Because C and M are fixed constants at this point, we may choose $N_2$ so that
$\color{green}{(C - 1)M}\frac{1}{n} < e/2 \tag{5}$ for all $n \ge N_2$. Finally, let $N$ [in $(1)$] $= \max\{C, N_2\}$ be the desired $N$.
4. Why are we authorized to choose $N_2$ so that $\color{green}{(C - 1)M}\frac{1}{n} < e/2$?
>$C \in \mathbb{N} \quad \therefore n - C < n \iff \color{magenta}{\frac{n - C}{n}} < 1 \tag{6}$
>$\begin{align}
\text{Equation (4) is} & & \color{green}{(C - 1)M}\frac{1}{n} & + \frac{e}{2}\color{magenta}{\frac{n - C}{n}}. \\
\text{Then by (5) and (6),} & & < e/2 & + e/2\color{magenta}{(1)}/ QED. \\
\end{align}$
[Stephen Abbott](http://www.middlebury.edu/academics/math/faculty/node/26391). [*Understanding Analysis* (2016 2 edn)](https://www.amazon.com/Understanding-Analysis-Stephen-Abbott/dp/1493927116). p. 55.
[1]: http://i.stack.imgur.com/VGDbn.png