Inequality possibly equal to Cauchy criterion?

Question

Let $a$ be a sequence.
I want to prove that there exists a number $n_0$ so that for all $x \geq n_0$ and $y \geq 3n_0$ $$|a_x - a_y| <\frac{1}{x\cdot\sqrt{y}}$$ implies that a converges.
I think that I would have to prove by construction by choosing $x$ and $y$ depending on a real number $\epsilon >0$ so that $\frac{1}{x\cdot\sqrt{y}} = \epsilon$ but I'm not sure how.
If this is the right idea, how do I find the right $x$ and $y$?
If not, what else can I do?

Answer 1

For any $p\ge1$, let $y=n+p,x=n$ and then the condition becomes $$ |a_{n+p}-a_n|<\frac{1}{n\sqrt{n+p}}<\frac{1}{n^{3/2}}. $$ For $\forall \epsilon>0$, let $N=\max\{3n_0,\frac{1}{\epsilon^{2/3}}\}$. Then if $n\ge N$, $$ |a_{n+p}-a_n|<\epsilon, \forall p\ge1 $$ which implies that $\{a_n\}$ is Cauchy. Therefore it converges.