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 : $$\sqrt{5}+1,\sqrt{5}-1,\sqrt{2}-1,\sqrt{2}+1$$


2 Answers 2


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 $$ \lim_{n\to\infty}\frac1n\sum_{r=1}^{2n}\frac{r}{\sqrt{n^2+r^2}}=\sqrt5-1\tag{7} $$

  • $\begingroup$ To tell the truth, I am a 12th standard student. The question is also from a high school book. I do not know Riemann Sum. Any simpler method will be preferred. $\endgroup$ Jun 12, 2014 at 3:18
  • 1
    $\begingroup$ @ArghyaChakraborty Doing series without calculus? How does that work? $\endgroup$
    – Shahar
    Jun 12, 2014 at 3:19
  • $\begingroup$ There must be another technique. The question is from a high school book, my knowledge is also of that level. $\endgroup$ Jun 12, 2014 at 3:21
  • 1
    $\begingroup$ Ok, What "high school book" is that and how come can they ask such a question without doing some basic calculus at Riemann Integrals level? I also see there "Cesaro sums"...but no integrals? Odd... $\endgroup$
    – DonAntonio
    Jun 12, 2014 at 3:40
  • 3
    $\begingroup$ In Australia most 12th year students also know how to calculate integrals but have not seen proper foundations such as Riemann sums, so I don't find the situation here so unbelievable. $\endgroup$ Jun 12, 2014 at 3:46

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 $\int^2_0 f(x) dx = \sqrt{5}-1.$


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