Question

${lim}_{n\to \infty }(\frac{n}{n^2}+\frac{n}{n^2+1}+\frac{n}{n^2+2^2}+......+\frac{n}{2n^2-2n+1})$ is equal to

• A. 1
• B. $\frac{\pi }{4}$
• C. tan 1
• D. $tan\frac{\pi }{4}$

Answer 1

Answer: (B)

$\frac{\pi }{4}$

${lim}_{n\to \infty }(\frac{n}{n^2}+\frac{n}{n^2+1}+\frac{n}{n^2+2^2}+......+\frac{n}{2n^2-2n+1})$
$={\sum }_{r=0}^{n-1}\frac{n}{n^2+r^2}=\frac{1}{n}={\sum }_{r=0}^{n-1}\frac{1}{1+\frac{r^2}{n^2}}$
Thus the given limit is equal to ${\int }_0^1\frac{dx}{1+x^2}=ta{n}^{-1}x{|}_0^1=\frac{\pi }{4}$
Ans: B