I suspect that $$ \lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^n\sum_{j=1}^n \frac{1}{\sqrt{i^2+j^2}} =a\approx 1.76$$ $$ \lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^n\sum_{j=1}^n \frac{1}{\sqrt{n^2+(i-j)^2}} =b\approx 0.934$$
Is there an analytic expression for $a$ or $b$?
You seem to have a typo in your value of $a$ - evaluating the sum as is yields something more like 1.76. Converting to an integral, $$ \lim_{n\rightarrow \infty } \frac{1}{n^2 } \sum_{i=1}^{n} \sum_{j=1}^{n} \frac{1}{\sqrt{(i/n)^2 + (j/n)^2 } } =\\ \int_{0}^{1} dx \int_{0}^{1} dy\, \frac{1}{\sqrt{x^2 +y^2 } } = 2\ln (1+\sqrt{2} ) $$ On the other hand, $$ \lim_{n\rightarrow \infty } \frac{1}{n^2 } \sum_{i=1}^{n} \sum_{j=1}^{n} \frac{1}{\sqrt{1+((i/n) - (j/n))^2 } } =\\ \int_{0}^{1} dx \int_{0}^{1} dy\, \frac{1}{\sqrt{1+(x-y)^2 } } = 2\ln (1+\sqrt{2} ) +2-2\sqrt{2} $$
