JEE Advanced · Mathematics · 27. Definite Integration
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
- 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 & Solution
Correct Answer
(D)
\(T_n>\frac{\pi}{3 \sqrt{3}}\)
Step-by-step Solution
Detailed explanation
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, \(T_n>\frac{\pi}{3 \sqrt{3}}\).
\[
\begin{aligned}
S_n & =\sum_{k=0}^n \frac{n}{n^2+k n+k^2} \\
& =\sum_{k=0}^n \frac{1}{n} \cdot\left(\frac{1}{1+\frac{k}{n}+\frac{k^2}{n^2}}\right) < \lim _{n \rightarrow \infty} \sum_{k=0}^n \frac{1}{n}\left(\frac{1}{1+\frac{k}{n}+\left(\frac{k}{n}\right)^2}\right) \\
& =\int_0^1 \frac{1}{1+x+x^2} d x=\left[\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2}{\sqrt{3}}\left(x+\frac{1}{2}\right)\right)\right]_0^1 \\
& =\frac{2}{\sqrt{3}} \cdot\left(\frac{\pi}{3}-\frac{\pi}{6}\right)=\frac{\pi}{3 \sqrt{3}}
\end{aligned}
\]
i.e., \(\quad S_n < \frac{\pi}{3 \sqrt{3}}\)
Similarly, \(T_n>\frac{\pi}{3 \sqrt{3}}\).
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