Question

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

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

Answer 1

Answer: (B)

Let I = $li{m}_{n\to \infty }\frac{1}{n}\sum _{r=1}^{2n}\frac{r}{\sqrt{n^2+r^2}}=li{m}_{n\to \infty }\frac{1}{n}\sum _{r=1}^{2n}\frac{r}{n\sqrt{1+(r/n{)}^2}}$
= $li{m}_{n\to \infty }\frac{1}{n}\sum _{r=1}^{2n}\frac{r/n}{\sqrt{1+(r/n{)}^2}}$
= $\underset{0}{\overset{2}{\int }}\frac{x}{\sqrt{1+x^2}}dx=[\sqrt{1+x^2}{]}_0^2=\sqrt{5}-1$