Show that a sequence is a Cauchy sequence

Question

Suppose that $(x_n)$ is a decreasing sequence of non-negative real numbers that converges to 0. Prove, using the definition, that the sequence $(y_n)$ where $y_n = x_1 - x_2 + x_3 - x_4 + \cdots + (-1)^{n{+1}}x_n$ is a Cauchy sequence.

I understand that a Cauchy sequence is a sequence that, for all $\epsilon < 0$ there exists a $p \in \mathbb{N}$ such that for all $m,n \in p$ $|x_m-x_n|< \epsilon$. However, I'm confused as to how to actually go about proving something is a Cauchy sequence and how in this case the sequence is built up from a different sequence. Any hints or suggestions would be appreciated, thank you!

Answer 1

Take $\varepsilon>0$. If $n$ is large enough, you have $x_N<\varepsilon$. Pick such a $N$. Now, if $m,n\geqslant N$, you want to prove that $\lvert y_m-y_n\rvert<\varepsilon$. This is trivial if $m=n$. If $m>n$, then$$y_m-y_n=\begin{cases}x_{n+1}-x_{n+2}+\cdots+\pm x_m&\text{ if }n+1\text{ is odd}\\-x_{n+1}+x_{n+1}-\cdots+\pm x_m&\text{ otherwise.}\end{cases}$$But, if $n+1$ is odd,$$y_m-y_n=x_{n+1}-(x_{n+2}-x_{n+3})-\cdots<x_{n+1}<\varepsilon$$and$$y_m-y_n=(x_{n+1}-x_{n+2})+(x_{n+3}-x_{n+4})+\cdots>0.$$Therefore, $\lvert y_m-y_n\rvert<\varepsilon$. The case in which $n+1$ is even is similar.

Answer 2

We need to show that for any $\epsilon > 0$, there exists a positive integer p such that for all m, n $\geq p$, the absolute difference between y_m and y_n is less than $\epsilon$, i.e., $|y_m - y_n| < \epsilon$.

Given $\epsilon > 0$, since the sequence converges to 0, there exists a positive integer $N_1$ such that for all n $\geq N_1$, we have $|x_n - 0| < \frac{\epsilon}{2}$.

Let m and n be two arbitrary positive constants such that $m, n \geq N_1$. Without loss of generality, assume that $m > n$. We want to show that $|y_m - y_n| < \epsilon$.

First, let's express $y_m - y_n$:

\begin{align*} y_m - y_n & = (x_1 - x_2 + x_3 - x_4 + \ldots + (-1)^{m+1}x_m) - (x_1 - x_2 + x_3 - x_4 + \ldots + (-1)^{n+1}x_n) \\ & = ((x_1 - x_2) + (x_3 - x_4) + \ldots + (-1)^{m+1}x_m) - ((x_1 - x_2) + (x_3 - x_4) + \ldots + (-1)^{n+1}x_n) \\ & = ((x_1 - x_2) + (x_3 - x_4) + \ldots + (-1)^{n}x_{n+1} + (-1)^{n+1}x_m) - ((x_1 - x_2) + (x_3 - x_4) + \ldots + (-1)^{n+1}x_n). \end{align*}

Now, we can group terms and simplify:

\begin{align*} y_m - y_n & = ((x_1 - x_2) + (x_3 - x_4) + \ldots + (-1)^{n}x_{n+1} + (-1)^{n+1}x_m) - ((x_1 - x_2) + (x_3 - x_4) + \ldots + (-1)^{n+1}x_n) \\ = ((x_1 - x_2) + (x_3 - x_4) + \ldots + (-1)^{n}x_{n+1} + (-1)^{n+1}x_m) \\ & \quad - (-(x_1 - x_2) - (x_3 - x_4) - \ldots - (-1)^{n+1}x_n) \quad \text{(Distribute the negative sign)} \\ & = ((x_1 - x_2) + (x_3 - x_4) + \ldots + (-1)^{n}x_{n+1} + (-1)^{n+1}x_m) \\ & \quad + ((x_1 - x_2) + (x_3 - x_4) + \ldots + (-1)^{n+1}x_n) \quad \text{(Remove the negative sign)}. \end{align*}

The following equation yields from the above equation by tringle inequality:

\begin{align*} |y_m - y_n| & \leq |(x_1 - x_2) + (x_3 - x_4) + \ldots + (-1)^{n}x_{n+1} + (-1)^{n+1}x_m| \\ & \quad + |(x_1 - x_2) + (x_3 - x_4) + \ldots + (-1)^{n+1}x_n| \\ & \leq |(x_1 - x_2)| + |(x_3 - x_4)| + \ldots + |(-1)^{n}x_{n+1}| + |(-1)^{n+1}x_m| \\ & \quad + |(x_1 - x_2)| + |(x_3 - x_4)| + \ldots + |(-1)^{n+1}x_n| \\ & = 2\left(|(x_1 - x_2)| + |(x_3 - x_4)| + \ldots + |(-1)^{n}x_{n+1}|\right). \end{align*}

Then

$2\left(|(x_1 - x_2)| + |(x_3 - x_4)| + \ldots + |(-1)^{n}x_{n+1}|\right)$ $\leq 2\left(\frac{\epsilon}{2} + \frac{\epsilon}{2} + \ldots + \frac{\epsilon}{2}\right) = n\epsilon$.

So, we have shown that for any $\epsilon > 0$ and for all m, n $\geq N_1$, $|y_m - y_n| < n\epsilon$.

Now, choose $p = N_1$ so that for all m, n $\geq p$, we have $|y_m - y_n| < n\epsilon$. Since n can be arbitrarily large, as m and n approach infinity, $|y_m - y_n|$ approaches 0. This satisfies the definition of a Cauchy sequence.