Prove that $(n+\frac {(-1)^n}{n}) $ is not Cauchy

Question

Let $x_n=(n+\dfrac {(-1)^n}{n}) $

So, $|x_{2m}-x_{2n}|=|(2m+\dfrac {(-1)^{2m}}{2m})-(2n+\dfrac {(-1)^{2n}}{2n})|$

$=|2m-2n+\dfrac{1}{2m}-\dfrac{1}{2n}|$

$=|(m-n) (2-\dfrac{1}{2mn})|$

This is where I'm stuck. What $\varepsilon$ can I chose to show that it is not Cauchy? I there any other $m,n$ I can choose to make this easier?

Answer 1

Since $m,n\ge 1$, you know that $2-\frac{1}{2mn} \ge 2-\frac{1}{2} = \frac{3}{2}$. Thus $|x_{2n}-x_{2m}| \ge \frac{3}{2}|m-n|$. Can you take it from here?

Answer 2

$\mathbb R$ is complete, so it suffices to show that the sequence doesn't converge. Consider the subsequence $$ x_{2n} = 2n + \frac{(-1)^{2n}}{2n} = 2n + \frac1{2n}.$$ Let $x\in\mathbb R$ and choose $n$ so that $2n>x+1$. Then $$\left| x_{2n} - x\right| = 2n + \frac1{2n} - x > 2n - x > 1,$$ so that $x_{2n}$ does not converge to $x$. Because $x$ was arbitrary, we have a subsequence of $x_n$ which does not converge. It follows that $x_n$ does not converge.