Show $\{x_n\}$ where $x_n = \frac{n}{n+1}$ is Cauchy

Question

So we know this is Cauchy because the sequence converges to 1. Now, to show it is Cauchy: $$|x_j - x_k| \leq \frac{1}{n} \ \forall n \in \mathbb{N} \text{ given } j,k \geq m$$

We know $|x_k - x_k| = |\frac{j}{j+1}- \frac{k}{k+1}| \leq |\frac{j}{j+1}| + |\frac{k}{k+1}| \leq \frac{1}{n}$

So really we can say $\frac{j}{j+1},\frac{k}{k+1} < \frac{1}{2n}$ Now to find $m$ we need to solve for $j$ or $k$.

We have, $$\frac{k}{k+1} = \frac{1}{2n}$$

$$\implies k= \frac{1}{2n-1}$$ So let m = $\frac{1}{2n-1}$ then $k,j > \frac{1}{2n-1}$ and $0 < \frac{k}{k+1},\frac{j}{j+1} < \frac{1}{2n-1}$

$$\frac{1}{n} > \frac{k}{k+1}+\frac{j}{j+1} \geq |\frac{k}{k+1}| + |\frac{j}{j+1}| \geq |\frac{k}{k+1} - \frac{j}{j+1}|$$

So really I am asking for help because I want to know if my approach works or not. I know there are other ways to approach the problem and get a solution. Is this a valid solution? I am missing a step or am I approaching the problem in the wrong way.

Answer 1

Your approach has problems. For example the inequality $$|\frac{j}{j+1}| + |\frac{k}{k+1}| \leq \frac{1}{n}$$

is not valid for large values of $j$ and $k$

You may try the following approach. $$|\frac {m}{m+1} - \frac {n}{n+1}|= \frac {|m-n|}{(m+1)(n+1)}\le$$

$$\frac {m}{mn}=\frac {1}{n}$$

So you make the difference of terms less than $\epsilon$ if you choose $m$ and $n$ larger than $\frac {1}{\epsilon}$

Answer 2

I'm not sure about your approach. We would like to make $|x_j - x_k|$ arbitrarily close to zero. Although we have $|x_j - x_k| \leq |\frac{j}{j+1}| + |\frac{k}{k+1}|$, notice that we can't make this upper bound arbitrarily close to zero. The bound I think you want is $|x_j-x_k| \leq \frac{1}{j} + \frac{1}{k}$. Convince yourself that this is true. Now given $\epsilon > 0$, can you find a natural number $M$ such that this upper bound is less than $\epsilon$ for $j,k > M$?

Answer 3

Very simply,

$\begin{array}\\ |x_j - x_k| &= |\frac{j}{j+1}- \frac{k}{k+1}|\\ &= |(1-\frac{1}{j+1})- (1-\frac{1}{k+1})|\\ &= |-\frac{1}{j+1}+\frac{1}{k+1}|\\ &\le |\frac{1}{j+1}|+|\frac{1}{k+1}|\\ &\lt 2|\frac{1}{\min(j, k)}|\\ &\to 0 \qquad\text{as } \min(j, k) \to \infty\\ \end{array} $

In general, if $x_k \to L$ as $k \to \infty$, then, for any $c > 0$, there is an $n(c)$ such that $|x_n-L| < c$ for $n > n(c)$.

Therefore, for any $c > 0$, for $j, k > n(c/2)$

$\begin{array}\\ |x_k - x_j| &= |(x_k-L)-(x_j-L)|\\ &\le |(x_k-L)|+|(x_j-L)|\\ &\lt (c/2)+c/2)\\ &= c\\ \text{so}\\ |x_k - x_j| &\to 0\\ \end{array} $