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If $x_n:=\sqrt{n}$, show that $(x_n)$ satisfies $\lim|x_{n+1}-x_n|=0$, but that it is not a Cauchy sequence.

We see that $$|x_{n+1}-x_n|=|\sqrt{n+1}-\sqrt{n}|=|(\sqrt{n+1}-\sqrt{n})\cdot\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}|=\bigl|\frac{1}{\sqrt{n+1}+\sqrt{n}}\bigr|$$

This seems to simplify nicely, but I am not quite sure where to proceed.

Relevant definitions:

  1. $\lim(x_n) = x$ if and only if for $\varepsilon>0$, $\exists k\in\mathbb{N}$ such that for all $n\geq k$, $|x_n-x|<\varepsilon$.

  2. A sequence $(x_n)$ is a Cauchy sequence if for every $\varepsilon>0$, $\exists h\in\mathbb{N}$ such that for all $n,m\geq h$, $|x_n-x_m|<\varepsilon$.

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Take $\epsilon=\frac12$. For any $h\in \mathbb{N}$, choose $n=h$ and $m=4h$. Then $|x_m-x_n|=|\sqrt{4h}-\sqrt{h}|=|2\sqrt{h}-\sqrt{h}|=\sqrt{h} \ge 1$ which contradicts the definition of Cauchy sequence for above $\epsilon$.

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You have shown that $$|x_{n+1}-x_n|=\frac1{\sqrt{n+1}+\sqrt n}\ ,$$ and this clearly approaches $0$ because $\sqrt{n+1}$ and $\sqrt n$ both approach $\infty$. (There is no need to use the definition with $\epsilon$ etc unless you have been instructed to do so.)

The easy way to show that $x_n$ is not Cauchy is to note that any Cauchy sequence is convergent (in $\Bbb R$), and $x_n=\sqrt n$ is not convergent.

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Hint (not intended to be a complete solution):$$\left|\dfrac{1}{\sqrt{n+1}+\sqrt{n}} \right| \leq \left|\dfrac{1}{\sqrt{n}+\sqrt{n}}\right| = \left|\dfrac{1}{2\sqrt{n}}\right|\text{.}$$ Take $k := \left \lceil{\left(\dfrac{1}{2\epsilon}\right)^2}\right \rceil $.

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