Find the region $(x,y) \in R$ for which the following sequence converges

$$\lim_{n \to \infty} \; \;\left| e^n\frac{(\sqrt{y}-\sqrt{x})^{2n}}{x^n} \right| = 0$$

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.

I would like to note that I have not had much exposure to working with multi-variable sequences, so my other question is: what literature would you recommend to add to my toolbox in dealing with these types of problems?

Your support is greatly appreciated!


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    $\begingroup$ Implicit is that $x, y > 0$. Why isn't the region simply $\{(x,y): (\sqrt y - \sqrt x)^2 < x/e\}$?? This is easily solved for $y$, etc. Or do you mean uniform convergence (in which case we look at the ratio being uniformly less than one)? $\endgroup$ – David Handelman Jun 14 '15 at 18:31

I cannot suggest any particular book, but I can find the region $R$ for the specific question, assuming I interpreted it correctly. Most introductory books on calculus and analysis should contain sufficient tools to come up with the following solution.

This sequence is of a very special form. I would not call it a multivariable sequence, but a usual single variable sequence with two real parameters. (I assume that $x$ and $y$ are real numbers in your question.)

If we denote $f(x,y)=e(\sqrt y-\sqrt x)^2/x$, the question is when does $\lim_{n\to\infty}|f(x,y)^n|=0$. This is true if $|f(x,y)|<1$ and false if $|f(x,y)|\geq1$. The case $|f(x,y)|=1$ is more delicate if you want to find the existence and the exact value of the limit.

To make sense of both square roots and dividing by $x$, we need to assume $x>0$ and $y\geq0$. Then $f(x,y)\geq0$. Now $$ \begin{split} & |f(x,y)|<1 \\\iff& f(x,y)<1 \\\iff& e(\sqrt y-\sqrt x)^2/x<1 \\\iff& e(\sqrt{y/x}-1)^2<1 \\\iff& -1/\sqrt e<\sqrt{y/x}-1<1/\sqrt e \\\iff& (1-1/\sqrt e)^2<y/x<(1+1/\sqrt e)^2. \end{split} $$ Similarly, $|f(x,y)|>1$ is equivalent with $y/x\notin[(1-1/\sqrt e)^2,(1+1/\sqrt e)^2]$. The borderline case is $y/x=(1\pm1/\sqrt e)^2$. In the borderline case we obtain simply $f(x,y)=1$, so we have $$ \lim_{n \to \infty} \; \;\left| e^n\frac{(\sqrt{y}-\sqrt{x})^{2n}}{x^n} \right| = \begin{cases} 0, & (1-1/\sqrt e)^2<y/x<(1+1/\sqrt e)^2\\ 1, & y/x=(1\pm1/\sqrt e)^2\\ \infty, & y/x\notin[(1-1/\sqrt e)^2,(1+1/\sqrt e)^2] \end{cases} $$ under the assumption that $x>0$ and $y\geq0$. The limit is always infinite if $y=0$.


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