Q32 / 69 · JEE Advanced 2008 (Paper 1) · Mathematics
Let $S_n=\sum_{k=0}^n \frac{n}{n^2+k n+k^2}$ and $T_n=\sum_{k=0}^{n-1} \frac{n}{n^2+k n+k^2}$, for $n=1,2,3, \ldots$, then
- Correct answerA$S_n < \frac{\pi}{3 \sqrt{3}}$
- B$S_n>\frac{\pi}{3 \sqrt{3}}$
- C$T_n < \frac{\pi}{3 \sqrt{3}}$
- Correct answerD$T_n>\frac{\pi}{3 \sqrt{3}}$
Solution
Given,
$$
\begin{aligned}
S_n & =\sum_{k=0}^n \frac{n}{n^2+k n+k^2} \\
& =\sum_{k=0}^n \frac{1}{n} \cdot\left(\frac{1}{1+\frac{k}{n}+\frac{k^2}{n^2}}\right) < \lim _{n \rightarrow \infty} \sum_{k=0}^n \frac{1}{n}\left(\frac{1}{1+\frac{k}{n}+\left(\frac{k}{n}\right)^2}\right) \\
& =\int_0^1 \frac{1}{1+x+x^2} d x=\left[\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2}{\sqrt{3}}\left(x+\frac{1}{2}\right)\right)\right]_0^1 \\
& =\frac{2}{\sqrt{3}} \cdot\left(\frac{\pi}{3}-\frac{\pi}{6}\right)=\frac{\pi}{3 \sqrt{3}}
\end{aligned}
$$
i.e., $\quad S_n < \frac{\pi}{3 \sqrt{3}}$
Similarly, $T_n>\frac{\pi}{3 \sqrt{3}}$.
Hence, the correct answer is options: 'A', 'D'.