Question

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

• A. 1 + $\sqrt 5$
• B. 1 - $\sqrt 5$
• C. - 1 + $\sqrt 2$
• D. 1 + $\sqrt 2$

Answer 1

Answer: (B)

1 - $\sqrt 5$

Answer 2

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