Evaluate $\lim_{n \to \infty} \sum_{j=0}^{n} \sum_{i=0}^j \frac{i^2+j^2}{n^4+ijn^2}$

Question

I am asked to evaluate: $$\lim_{n \to \infty} \sum_{j=0}^{n} \sum_{i=0}^j \frac{i^2+j^2}{n^4+ijn^2}$$

I am not experienced with double summations, but I tried simplifying the expression above into: $$\lim_{n \to \infty}\frac{1}{n^2} \sum_{j=0}^{n} \sum_{i=0}^j \frac{{\left(\frac{i}{n}\right)}^2+{\left(\frac{j}{n}\right)}^2}{1+\left(\frac{i}{n}\right) \left(\frac{j}{n}\right)}$$ Perhaps this is Riemann integral (double)?

Answer 1

HINT: Your Riemann Integral idea was a good one. We can express your limit using a double integral as follows:

$$\int_0^1 \int_0^x \frac{x^2+y^2}{1+xy}\space dy \space dx$$

Can you take it from here?

Answer 2

To evaluate the integral we use symmetries and geometric and telescoping series viz.$$\begin{align}\int_0^1\int_0^x\frac{x^2+y^2}{1+xy}dydx&=\frac12\int_0^1\int_0^1\frac{x^2+y^2}{1+xy}dxdy\\&=\int_0^1\int_0^1\frac{x^2dxdy}{1+xy}\\&=\sum_{n\ge0}(-1)^n\int_0^1x^{n+2}dx\int_0^1y^ndy\\&=\sum_{n\ge0}\frac{(-1)^n}{(n+1)(n+3)}\\&=\frac12\sum_{n\ge0}\left(\frac{(-1)^n}{n+1}-\frac{(-1)^n}{n+3}\right)\\&=\frac12\left(1-\frac12\right)\\&=\frac14.\end{align}$$