How to prove $(-1)^n$ is not Cauchy in $\mathbb{R}$?

Question

I am an engineer and I would like to prove that the sequence $(-1)^n$ is not a Cauchy sequence, in order to understand the definition of Cauchy sequence better.

Thank you.

Answer 1

Think of it this way : The sequence $(-1)^n$ is really made up of two sequences $\{1,1,1,\ldots\}$ and $\{-1, -1, -1,\ldots\}$ which are both going in different directions.

A Cauchy sequence is, for all intents and purposes, a sequence which "should" converge (It may not, but for sequences of real numbers, it will). So, if $(-1)^n$ should converge, where will it converge to? $-1$ and $1$ are both reasonable answers, so the answer is "it cannot converge" :)

Answer 2

A sequence $x_{1},x_{2},\ldots$ is Cauchy if $\forall \epsilon$ $\exists N \in \mathbb{N}$ such that $m,n > N \Rightarrow |x_{m} - x_{n}| < \epsilon$. Or, in English, a sequence is Cauchy if after some point any pair of elements in the sequence are as close as you'd like to each other.

Can you find $N$ so that any pair of elements after $N$ is within a distance of $1$ from each other?

No. Whatever value you pick for $N$ the sequence after $x_{N}$ will contain some $x_{p} = 1$ and another $x_{q} = x_{p+1} = -1$. The distance between these two elements is $2$ which is (obviously) more than $1$.

Answer 3

To briefly recall the definition of a Cauchy sequence:

A sequence $\{x_n\}_{n=1}^{\infty}$ is said to be Cauchy if, given an $\epsilon > 0$ we have a $N \in \mathbb N$ such that for all $n,m > N$ we have that $|x_n - x_m| < \epsilon$

From this, think what it means for a sequence to not be Cauchy - once you've done that look below:

A sequence $\{x_n\}_{n=1}^{\infty}$ is not Cauchy if there exists an $\epsilon > 0$ such that for all $N \in \mathbb N$ such that we have a pair $n(N),m(N)$ where $n(N),m(N)>N$ such that $|x_n - x_m| \geq \epsilon$

Now, as we have that for the sequence defined by $x_n = (-1)^n$ that

$$|x_m - x_n| = \begin{cases} 0, & \text{if $m$ and $n$ have the same parity} \\ 2, & \text{if $m$ and $n$ do not have the same parity} \\ \end{cases}$$

where by parity we mean whether a number is odd or even, we see that if we were to choose $\epsilon = 1$ for example, then given any $N \in \mathbb N$ we can choose an even number $m$ and odd number $n$ bigger than $N$ such that $|x_m - x_n| = 2 > 1 = \epsilon$, and hence the sequence is not Cauchy. Of course, the $m$ and $n$ we choose are effectively arbitary, but if you wanted to go a bit deeper, you could specify values for $n,m$ - the easiest choice is to simply choose $N$ and $N+1$.