use the method in second image example 20.2 Prove that both of the following sequences are Cauchy sequences, using just the definition (not any Theorems):(a) $\left\{ \frac{(-1)^n}{n} \right\}_{n=1}^{\infty}$(b) $\left\{ \frac{n}{n+1} \right\}_{n=1}^{\infty}$ **Example 20.2.** I claim the sequence $\left\{\frac{1}{n}\right\}$ is a Cauchy sequence.**Proof.** Pick $\varepsilon > 0$. (Scratch work: we need to figure out how big $m$ and $n$ need to be in order that $\left|\frac{1}{n} - \frac{1}{m}\right| < \varepsilon$. Note that if both $\frac{1}{n}$ and $\frac{1}{m}$ are between 0 and $\varepsilon$, then the distance between them will be at most $\varepsilon$, and this occurs so long as $n, m > \frac{1}{\varepsilon}$. So we will set $N = \frac{1}{\varepsilon}$. Back to the proof.) Let $N = \frac{1}{\varepsilon}$. Let $m, n \in \mathbb{N}$ be such that $m > N$ and $n > N$. Then $0 < \frac{1}{n} < \frac{1}{N} = \varepsilon$ and $0 < \frac{1}{m} < \frac{1}{N} = \varepsilon$. It follows that\[|a_m - a_n| = \left|\frac{1}{m} - \frac{1}{n}\right| \leq \varepsilon\]and this proves the sequence is Cauchy. □

(a) Consider the sequence, an=-1nnn=1∞ Let ε>0, For, n,m∈ℕ, if n≥m then 1n≤1m Then,…

Answered: Prove that both of the following…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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use the method in second image example 20.2

Prove that both of the following sequences are Cauchy sequences, using just the definition (not any Theorems):

(a) $\left\{ \frac{(-1)^n}{n} \right\}_{n=1}^{\infty}$

(b) $\left\{ \frac{n}{n+1} \right\}_{n=1}^{\infty}$

**Example 20.2.** I claim the sequence $\left\{\frac{1}{n}\right\}$ is a Cauchy sequence.

**Proof.** Pick $\varepsilon > 0$. (Scratch work: we need to figure out how big $m$ and $n$ need to be in order that $\left|\frac{1}{n} - \frac{1}{m}\right| < \varepsilon$. Note that if both $\frac{1}{n}$ and $\frac{1}{m}$ are between 0 and $\varepsilon$, then the distance between them will be at most $\varepsilon$, and this occurs so long as $n, m > \frac{1}{\varepsilon}$. So we will set $N = \frac{1}{\varepsilon}$. Back to the proof.) Let $N = \frac{1}{\varepsilon}$. Let $m, n \in \mathbb{N}$ be such that $m > N$ and $n > N$. Then $0 < \frac{1}{n} < \frac{1}{N} = \varepsilon$ and $0 < \frac{1}{m} < \frac{1}{N} = \varepsilon$. It follows that

\[
|a_m - a_n| = \left|\frac{1}{m} - \frac{1}{n}\right| \leq \varepsilon
\]

and this proves the sequence is Cauchy. □

Expert Solution

Step 1

(a)

Consider the sequence, $\{a_{n}\} = {\{\frac{{(- 1)}^{n}}{n}\}}_{n = 1}^{\infty}$

Let $ε > 0$ ,

For, $n, m \in ℕ$ , if $n \geq m$ then $\frac{1}{n} \leq \frac{1}{m}$

Then,

$\begin{array}{rcl} |a_{n} - a_{m}| & = & |\frac{{(- 1)}^{n}}{n} - \frac{{(- 1)}^{m}}{m}| \\ \leq & |\frac{{(- 1)}^{n}}{n}| + |\frac{{(- 1)}^{m}}{m}| \\ \leq & \frac{1}{n} + \frac{1}{m} \\ \leq & \frac{1}{m} + \frac{1}{m} = \frac{2}{m} < ε \end{array}$

if $m > \frac{2}{ε}$

Let, $N$ be a positive integer, such that, $N \geq \frac{2}{ε}$

Then, $\begin{array}{rcl} |a_{n} - a_{m}| \end{array} < ε$ , for $n \geq m$ .

So, $\{a_{n}\} = {\{\frac{{(- 1)}^{n}}{n}\}}_{n = 1}^{\infty}$ is a Cauchy sequence.

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