Sum-it-up!

Since $\begin{aligned} \frac{1}{n^{\frac52}}\sum_{i=1}^{n^2-1}\sqrt{n + \sqrt{i}} & = \; \frac{1}{n^2} \sum_{i=1}^{n^2-1}\sqrt{1 + \sqrt{\tfrac{i}{n^2}}} \; \to \; \int_0^1 \sqrt{1 + \sqrt{x}}\,dx \\ \frac{1}{n^{\frac52}}\sum_{i=1}^{n^2-1}\sqrt{n - \sqrt{i}} & = \; \frac{1}{n^2} \sum_{i=1}^{n^2-1}\sqrt{1 - \sqrt{\tfrac{i}{n^2}}} \; \to \; \int_0^1 \sqrt{1 - \sqrt{x}}\,dx \end{aligned}$ as $n \to \infty$ , and since elementary integration gives $\int_0^1 \sqrt{1 + \sqrt{x}}\,dx \; = \; \tfrac{8}{15}(1 + \sqrt{2}) \hspace{2cm} \int_0^1 \sqrt{1 - \sqrt{x}}\,dx \; = \; \tfrac{8}{15}$ we deduce that $\lim_{n \to \infty}a_n \; = \; \frac{\displaystyle \int_0^1 \sqrt{1 + \sqrt{x}}\,dx}{\displaystyle \int_0^1 \sqrt{1 - \sqrt{x}}\,dx} \; = \; \boxed{1 + \sqrt{2}}$

Now , we observe that the fraction $\frac{"x"}{"y"}$ does not depend on the value of n. Thus, the answer will remain same even if n approaches infinity ..