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

Answer: (A), (D)

Given, $S_{n}=\displaystyle\sum_{k=0}^{n} \frac{n}{n^{2}+k n+k^{2}}$
$=\displaystyle\sum_{k=0}^{n} \frac{1}{n} \cdot\left(\frac{1}{1+\frac{k}{n}+\frac{k^{2}}{n^{2}}}\right) < \displaystyle\lim _{n \rightarrow \infty} \displaystyle\sum_{k=0}^{n} \frac{1}{n}\left(\frac{1}{1+\frac{k}{n}+\left(\frac{k}{n}\right)^{2}}\right)$
$=\int\limits_{0}^{1} \frac{1}{1+x+x^{2}} d x=\left[\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2}{\sqrt{3}}\left(x+\frac{1}{2}\right)\right)\right]_{0}^{1}$
$=\frac{2}{\sqrt{3}} \cdot\left(\frac{\pi}{3}-\frac{\pi}{6}\right)=\frac{\pi}{3 \sqrt{3}}$
i.e. $ S_{n} < \frac{\pi}{3 \sqrt{3}}$
Similarly, $ T_{n}>\frac{\pi}{3 \sqrt{3}}$