Evaluate $\mathrm{lim}_{n \to\infty } n^{-2} \sum_{i=1}^{n} \sum_{j=1}^{n^2} \frac{1}{\sqrt{n^2 + ni + j}}$. I really don't have much idea for this one. I've recognised that it's very similar to a double integral of a rectangular region R that is broken up into $n^3$ smaller areas.
The rectangle have dimensions $a \text{ x } a^2$ for some $a$.
If the rectangle is broken up into $n^3$ rectangles that are $\frac{a}{n} \text{ x } \frac{a}{n}$, then each has $\Delta A = \frac{a^2}{n^2}$.
Hence if a = 1, 
$\mathrm{lim}_{n \to\infty } n^{-2} \sum_{i=1}^{n} \sum_{j=1}^{n^2} \frac{1}{\sqrt{n^2 + ni + j}}$ = $\mathrm{lim}_{n \to\infty } \sum_{i=1}^{n} \sum_{j=1}^{n^2} \frac{1}{\sqrt{n^2 + ni + j}}\Delta A$.
I don't exactly have a reason for choosing a = 1 other than it fit best.
And from there I am out of ideas. I haven't really come across a problem like this before.
 A: Hint:
$$\frac{1}{{{n^2}}}\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^{{n^2}} {\frac{1}{{\sqrt {{n^2} + ni + j} }}} }  = \frac{1}{{{n^3}}}\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^{{n^2}} {\frac{1}{{\sqrt {1 + \frac{i}{n} + \frac{j}{{{n^2}}}} }}} }  \to \int_0^1 {\int_0^1 {\frac{1}{{\sqrt {1 + x + y} }}dxdy} } $$

If this is not clear to you, consider the Riemann sum corresponding to partition points $(i/n,j/n^2)$ with $i=1,2,..,n$, $j=1,2,...,n^2$.
A: And just to complete pisco125's answer,
$$\begin{eqnarray*} \iint_{(0,1)^2}\frac{dx\,dy}{\sqrt{1+x+y}} &\stackrel{\text{symmetry}}{=}&2\iint_{0\leq y\leq x\leq 1}\frac{dx\,dy}{\sqrt{1+x+y}}\\&=&2\int_{0}^{1}\int_{0}^{x}\frac{1}{\sqrt{1+x+y}}\,dy\,dx\\&\stackrel{y\to x u}{=}& 2\int_{0}^{1}\int_{0}^{1}\frac{x}{\sqrt{1+x+xu}}\,du\,dx\\&=&4\int_{0}^{1}\sqrt{1+2x}-\sqrt{1+x}\,dx\\&=&\frac{4}{3}\left[(1+2x)^{3/2}-2(1+x)^{3/2}\right]_{0}^{1}\\&=&\color{blue}{\frac{4}{3}\left(3\sqrt{3}-4\sqrt{2}+1\right)}\approx\frac{215}{299}. \end{eqnarray*}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\lim_{n \to \infty}n^{-2}\sum_{i = 1}^{n}
\sum_{j = 1}^{n^{2}}{1 \over \sqrt{n^{2} + ni + j}}} =
\lim_{n \to \infty}n^{-2}\sum_{i = 1}^{n}
\sum_{j = n^{2} + ni + 1}^{2n^{2} + ni}{1 \over \sqrt{j}}
\\[5mm] = &\
\lim_{n \to \infty}n^{-2}\sum_{i = 1}^{n}
\pars{\sum_{j = 1}^{2n^{2} + ni}{1 \over \sqrt{j}} -
\sum_{j = 1}^{n^{2} + ni}{1 \over \sqrt{j}}}
\\[5mm] = &\
2\lim_{n \to \infty}n^{-2}\sum_{i = 1}^{n}
\pars{\root{2n^{2} + ni} -
\root{n^{2} + ni}}
\\[5mm] = &\
2\lim_{n \to \infty}\sum_{i = 1}^{n}
{1 \over \root{2n^{2} + ni} + \root{n^{2} + ni}}
\\[5mm] = &\
2\lim_{n \to \infty}{1 \over n}\sum_{i = 1}^{n}
{1 \over \root{2 + i/n} + \root{1 + i/n}} =
2\int_{0}^{1}{\dd x \over \root{x + 2} + \root{x + 1}}
\\[5mm] = &\
\bbx{4\root{3} - {16 \over 3}\root{2} + {4 \over 3}}
\approx 0.7191 \\ &
\end{align}
