Question

$n\stackrel{Lt}{\to }\infty \sum _{r=0}^{n-1}\frac{1}{\sqrt{n^2-r^2}}$

• A. $\pi$
• B. $\pi /2$
• C. $\pi /3$
• D. $\pi /6$

Answer 1

The correct option is B $\pi /2$

$n\stackrel{Lt}{\to }\infty {\sum }_{r=0}^{n-1}\frac{1}{\sqrt{n^2-r^2}}=n\stackrel{Lt}{\to }\infty {\sum }_{r=0}^n\frac{1}{n}\cdot \frac{1}{\sqrt{1-(\frac{r}{n}{)}^2}}$
It is in the form of
$n\stackrel{Lt}{\to }\infty {\sum }_{r=0}^n\frac{1}{n}f\left(\frac{r}{n}\right)={\int }_0^{1}f\left(x\right)dx$
${\int }_0^1\frac{1}{\sqrt{1-x^2}}dx$
$=\left[si{n}^{-1}\right(x)+c]{\int }_0^1$
$=si{n}^{-1}\left(1\right)=\frac{\pi }{2}$