Question

Let $S_n= \displaystyle \sum_{k=0}^n \frac{n}{n^2+kn+k^2} \, and \, T_n= \displaystyle \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

The Correct Option is D

Given, Sn = $ \displaystyle \sum_{k=0}^n \frac{n}{n^2+kn+k^2}$
$=\displaystyle \sum_{k=0}^n \frac{1}{n}.\Bigg(\frac{1}{1+\frac{k}{n}+\frac{k^2}{n^2}}\Bigg) \, \, < lim_{n \to \, \infty} \, \displaystyle \sum_{k=0}^n \frac{1}{n}.\Bigg(\frac{1}{1+\frac{k}{n}+\bigg(\frac{k}{n}\bigg)^2}\Bigg)$
$\int_0^1 \frac{1}{1+x+x^2}dx = \bigg[ \frac{2}{\sqrt 3} tan^{-1} \bigg(\frac{2}{\sqrt 3}\bigg(x+\frac{1}{2}\bigg)\bigg)\bigg]_0^1$
$=\frac{2}{\sqrt 3}.\bigg(\frac{\pi}{3}-\frac{\pi}{6}\bigg) =\frac{\pi}{3\sqrt 3}$
$i.e.., \, \, \, \, \, \, \, \, \, \, \, \, S_n < \frac{\pi}{3\sqrt 3}$
Similarly, $ \, \, \, \, \, \, \, \, \, \, T_n > \frac{\pi}{3\sqrt 3}$