Question

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

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

Answer 1

The correct option is B $-1+\sqrt{5}$

$\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^{2n}\frac{r}{\sqrt{n^2+r^2}}=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^{2n}\frac{\frac{r}{n}}{\sqrt{1+\frac{r^2}{n^2}}}$
$={\int }_0^2\frac{x}{\sqrt{1+x^2}}dx={\left[\sqrt{1+x^2}\right]}_0^2=\sqrt{5}-1$