Question

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

The correct option is A $S_n<\frac{\pi }{3\sqrt{3}}$

$\begin{array}{c}Let,f\left(x\right)=\frac{1}{x^2+x+1}\\ so,S_n<\underset{0}{\overset{1}{\int }}f\left(x\right)<T_n\\ \underset{0}{\overset{1}{\int }}\begin{array}{c}f\left(x\right)=\underset{0}{\overset{1}{\int }}\left(\frac{1}{{\left(x+\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\right)dx\\ =\frac{2}{\sqrt{3}}{\tan }^{-1}{\left(\frac{2x+1}{\sqrt{3}}\right)|}_0^1\\ =\frac{2}{\sqrt{3}}\left[\frac{\pi }{3}-\frac{\pi }{6}\right]=\frac{\pi }{3\sqrt{3}}\\ S_n<\frac{\pi }{3\sqrt{3}}<T_n\end{array}\end{array}$