I'm given a sequence of real nos. satisfying the condition$$|x_{n+1} - x_n| = \frac{1}{\sqrt{n}},\,\,\forall n \in \mathbb N$$
I'm trying to prove that a sequence satisfying this condition will never be a Cauchy sequence.
My attempt:I tried constructing $|x_{n+p} - x_n|$,
$$|x_{n+p}-x_n|= |x_{n+p}-x_{n+p-1}+x_{n+p-1}-x_{n+p-2}+x_{n+p-2}-\cdots+x_{n+1}-x_n|$$$$\therefore |x_{n+p}-x_n|\le |x_{n+p}-x_{n+p-1}|+|x_{n+p-1}-x_{n+p-2}|+|x_{n+p-2}- x_{n+p-3}|+\cdots+|x_{n+1}-x_n|$$$$ |x_{n+p}-x_n|\le \frac{1}{\sqrt{n+p-1}}+\frac{1}{\sqrt{n+p-2}}+\cdots+\frac{1}{\sqrt{n}}$$
Next, I tried creating a lower bound for the sum when $p = n+1$$$\frac{1}{\sqrt{2n}}+\frac{1}{\sqrt{2n}}+\cdots n+1\,terms\le\frac{1}{\sqrt{n+(n+1)-1}}+\frac{1}{\sqrt{n+(n+1)-2}}+\cdots+\frac{1}{\sqrt{n}}$$$$\frac{1}{\sqrt{2}}\lt\frac{n+1}{\sqrt{2n}}\le\frac{1}{\sqrt{n+(n+1)-1}}+\frac{1}{\sqrt{n+(n+1)-2}}+\cdots+\frac{1}{\sqrt{n}}$$
I can't proceed beyond this point as I can't extend this lower bound to $|x_{2n+1} - x_n|$.
I'm also unable to come up with an example sequence that would satisfy this condition and is Cauchy.