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Find $$\lim_{n \to \infty} \frac{1}{n}\sum^{2n}_{r =1} \frac{r}{\sqrt{n^2+r^2}}$$

My approach :

$$\lim_{n \to \infty} \frac{1}{n}\sum^{2n}_{r =1} \frac{r}{\sqrt{n^2+r^2}} =\lim_{n \to \infty} \frac{1}{n^2}\sum^{2n}_{r =1} \frac{\frac{r}{n}}{\sqrt{1+\frac{r^2}{n^2}}} $$

If I put $\frac{r}{n} =t $ then we can write it

$$\lim_{n \to \infty} \frac{1}{n^2}\sum^{2n}_{r =1} \frac{t}{\sqrt{1+t^2}} $$ Will it help some how here.. and how can we change the limits then.. please suggest thanks.

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  • $\begingroup$ do you know how to evaluate a sum using definite integrals? $\endgroup$ – dajoker Aug 11 '13 at 15:32
  • $\begingroup$ Some MathJax advice: Named math operators should appear upright, and the common ones have their own code for this purpose (e.g. \sin, \log - see entry 11 in our MathJax guide). $\endgroup$ – Zev Chonoles Aug 11 '13 at 15:39
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HINT:

Putting $2n=m,$

$$\lim_{n \to \infty} \frac{1}{n}\sum^{2n}_{r =1} \frac{r}{\sqrt{n^2+r^2}} $$

$$=4\lim_{n \to \infty} \frac{1}{2n}\sum^{2n}_{r =1} \frac{r}{\sqrt{(2n)^2+4r^2}} $$

$$=4\lim_{m\to\infty}\frac1m\sum^m_{r=1}\frac r{\sqrt{m^2+4r^2}}$$

$$=4\lim_{m\to\infty}\frac1m\sum^m_{r=1}\frac {\frac rm}{\sqrt{1+4\left(\frac rm\right)^2}}$$

$$=4\int_0^1\frac{xdx}{\sqrt{1+4x^2}}$$

as $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$

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