Question

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

• (A) $s_n<\frac{\pi }{3\sqrt{3}}$
• (B) $S_n>\frac{\pi }{3\sqrt{3}}$
• (C) ${\mathrm{T}}_n<\frac{\pi }{3\sqrt{3}}$
• (D) ${\mathrm{T}}_n>\frac{\pi }{3\sqrt{3}}$

Answer 1

The Correct Option is A, D

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