Question

Let $S_n=\sum _{k=0}^n\frac{n}{n^2+kn+k^2}$ and $T_n=\sum _{k=0}^{n-1}\frac{1}{n^2+kn+k^2},for$
$n=1,2,3$,... Then,

• A. $S_n<\frac{\pi }{3\sqrt{3}}$
• B. $S_n>\frac{\pi }{3\sqrt{3}}$
• C. $T_n<\frac{\pi }{3\sqrt{3}}$
• D. $T_n>\frac{\pi }{3\sqrt{3}}$

Answer 1

Answer: (A), (D)

Given, $S_n=\sum _{k=0}^n\frac{n}{n^2+kn+k^2}$
$=\sum _{k=0}^n\frac{1}{n}\cdot \left(\frac{1}{1+\frac{k}{n}+\frac{k^2}{n^2}}\right)<\underset{n\to \infty }{\lim }\sum _{k=0}^n\frac{1}{n}\left(\frac{1}{1+\frac{k}{n}+{\left(\frac{k}{n}\right)}^2}\right)$
$=\underset{0}{\overset{1}{\int }}\frac{1}{1+x+x^2}dx={\left[\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2}{\sqrt{3}}\left(x+\frac{1}{2}\right)\right)\right]}_0^1$
$=\frac{2}{\sqrt{3}}\cdot \left(\frac{\pi }{3}-\frac{\pi }{6}\right)=\frac{\pi }{3\sqrt{3}}$
i.e. $S_n<\frac{\pi }{3\sqrt{3}}$
Similarly, $T_n>\frac{\pi }{3\sqrt{3}}$