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Q. $ lim_{ n \to \infty} \frac{1}{n} \displaystyle \sum_{r = 1}^ {2n} \frac{r}{ \sqrt{ n^2 + r^2}} $ equals

IIT JEEIIT JEE 1999

Solution:

Let I = $ lim_{ n \to \infty} \frac{1}{n} \displaystyle \sum_{r = 1}^ {2n} \frac{r}{ \sqrt{ n^2 + r^2}} = lim_{n \to \infty} \frac{1}{n} \displaystyle \sum_{r = 1}^ {2n} \frac{r}{ n \sqrt{ 1 + (r / n)^2}} $
= $ lim_{ n \to \infty} \frac{1}{n} \displaystyle \sum_{r = 1}^ {2n} \frac{ r / n}{ \sqrt {1 + ( r / n)^2}} $
= $ \int \limits_0^2 \frac{x}{\sqrt{ 1 + x^2}} dx = [ \sqrt{ 1 + x^2 }]_0^2 = \sqrt 5 - 1 $