Definite Integration Question 5
Question 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, $\quad T_{n}>\frac{\pi}{3 \sqrt{3}}$
