Value of the given limit I need to calculate the value of :
$$\lim_{n\to \infty}\frac{1}{n}\sum_{r=1}^{2n}{\frac{r}{\sqrt{n^2+r^2}}}$$
I had been trying to use Cesàro summation but somehow, I might be messing up.
The options are : $$\sqrt{5}+1,\sqrt{5}-1,\sqrt{2}-1,\sqrt{2}+1$$
 A: Hint: As a Riemann Sum:
$$
\lim_{n\to\infty}\sum_{r=1}^{2n}\frac{r/n}{\sqrt{1+r^2/n^2}}\frac1n=\int_0^2\frac{x}{\sqrt{1+x^2}}\mathrm{d}x\tag{1}
$$

We can also use the fact that
$$
\sqrt{n^2+r^2}-\sqrt{n^2+(r-1)^2}=\frac{2r-1}{\sqrt{n^2+r^2}+\sqrt{n^2+(r-1)^2}}\tag{2}
$$
and
$$
\begin{align}
&\left|\,\frac{2r-1}{\sqrt{n^2+r^2}+\sqrt{n^2+(r-1)^2}}-\frac{r}{\sqrt{n^2+r^2}}\,\right|\\[6pt]
&=\small\left|\,\frac{r}{\sqrt{n^2+r^2}}\frac{\sqrt{n^2+r^2}-\sqrt{n^2+(r-1)^2}}{\sqrt{n^2+r^2}+\sqrt{n^2+(r-1)^2}}-\frac1{\sqrt{n^2+r^2}+\sqrt{n^2+(r-1)^2}}\,\right|\\[6pt]
&=\small\left|\,\frac{r}{\sqrt{n^2+r^2}}\frac{2r-1}{\left(\sqrt{n^2+r^2}+\sqrt{n^2+(r-1)^2}\right)^2}-\frac1{\sqrt{n^2+r^2}+\sqrt{n^2+(r-1)^2}}\,\right|\\[6pt]
&\le\frac{2n}{4n^2}+\frac1{2n}\\[12pt]
&\le\frac1n\tag{3}
\end{align}
$$
Therefore, using $(3)$ gives
$$
\left|\,\frac1n\sum_{r=1}^{2n}\left[\frac{2r-1}{\sqrt{n^2+r^2}+\sqrt{n^2+(r-1)^2}}-\frac{r}{\sqrt{n^2+r^2}}\right]\,\right|\le\frac2n\tag{4}
$$
and using $(2)$ yields
$$
\frac1n\sum_{r=1}^{2n}\frac{2r-1}{\sqrt{n^2+r^2}+\sqrt{n^2+(r-1)^2}}=\sqrt5-1\tag{5}
$$
Finally, $(4)$ and $(5)$ say
$$
\left|\,\frac1n\sum_{r=1}^{2n}\frac{r}{\sqrt{n^2+r^2}}-(\sqrt5-1)\,\right|\le\frac2n\tag{6}
$$
Therefore, $(6)$ gives
$$
\lim_{n\to\infty}\frac1n\sum_{r=1}^{2n}\frac{r}{\sqrt{n^2+r^2}}=\sqrt5-1\tag{7}
$$
A: Draw a sketch of $f(x)=\dfrac{x}{\sqrt{1+x^2}}$ between $x=0$ and $x=2.$ Calculate the derivative to see the function is increasing in this interval.
Mark the points $0 \ , \ 1/n \ , \ 2/n \ , \ \ldots \ , \ 2$ on the $x$-axis. Now on each interval $\left[  \ \dfrac{i-1}{n} \ , \ \dfrac{i}{n} \right]$ draw a rectangle of height $f(i/n).$ Notice that the sum of the areas of these rectangles is $S_n := \displaystyle \frac{1}{n}\sum_{r=1}^{2n}{\frac{r}{\sqrt{n^2+r^2}}},$ and that this overestimates the area $\int^2_0 f(x) dx.$ So we get $S_n \geq \int^2_0 f(x) dx.$ 
Now do the same thing, but instead of drawing the rectangle with height $f(i/n),$ take the height as $f((i-1)/n).$ Then these rectangles underestimate the area, and the sum of the areas of the rectangles is $S_n - \dfrac{2}{\sqrt{5} n}.$ So you have 
$$ S_n - \frac{2}{\sqrt{5} n } \leq \int^2_0 f(x) dx \leq S_n.$$
Taking the limit as $n\to \infty$ you see that your limit is equal to $\int^2_0 f(x) dx = \sqrt{5}-1.$  
