Indefinite Integration 4 Question 5
5. 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) $T _n>\frac{\pi}{3 \sqrt{3}}$
$(2008,4$ M)
Analytical & Descriptive Question
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Answer:
Correct Answer: 5. (b, d)
Solution:
- Given, $S _n=\sum _{k=0}^{n} \frac{n}{n^{2}+k n+k^{2}}$
$$ \begin{aligned} & =\sum _{k=0}^{n} \frac{1}{n} \cdot \frac{1}{1+\frac{k}{n}+\frac{k^{2}}{n^{2}}}<\lim _{n \rightarrow \infty} \sum _{k=0}^{n} \frac{1}{n} \frac{1}{1+\frac{k}{n}+\frac{k^{2}}{n}} \\ & =\int _0^{1} \frac{1}{1+x+x^{2}} d x \\ & =\frac{2}{\sqrt{3}} \tan ^{-1} \frac{2}{\sqrt{3}} x+\frac{1}{2} \\ & =\frac{2}{\sqrt{3}} \cdot \frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{3 \sqrt{3}} \quad \text { i.e. } S _n<\frac{\pi}{3 \sqrt{3}} \end{aligned} $$
Similarly, $\quad T _n>\frac{\pi}{3 \sqrt{3}}$
