Find the region $(x,y) \in R$ for which the following sequence converges
$$\left| e^n\frac{(\sqrt{y}-\sqrt{x})^{2n}}{x^n} \right| \to 0$$
I am currently doing number theory research on studying the irrational numbers. As I am working on this, I find myself stumped in determining the region $R$. Aside from the trivial solution (e.g. $x=y$), I'm not sure how exactly I can extract $R$ from the sequence.
In addition to help with the problem, what literature would you recommend to add to my toolbox in dealing with these types of problems?
Thank you for your support!
I'm assuming you're only working with real numbers. In that case, to be taking square roots you have to have $x > 0$ and $y \geq 0$. To solve this problem, let's fix $x$, and then find all $y$ such that the given thing converges.
So given $x$, let $z = \sqrt{x}$, so $$e^n \frac{(\sqrt{y} - \sqrt{x})^{2n}}{x^n} = \sqrt{e}^{2n} \frac{(\sqrt{y} - z)^{2n}}{z^{2n}}$$ This expression converges to $0$ as $n$ goes to infinity if and only if its square root $$\frac{\sqrt{e}^n|\sqrt{y} - z|^n}{z^n}$$ also converges to $0$ as $n$ tends to infinity. Now, when does $\frac{a^n}{b^n}$ tend to $0$? If and only if $a < b$.
So $\sqrt{e}|\sqrt{y} - z| < z$, or $|\sqrt{y} - \sqrt{x}|^2 < \frac{x}{e}$.
So your answer is whatever the graph of $|\sqrt{y} - \sqrt{x}|^2 < \frac{x}{e}$ looks like.