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)

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Answer:

Correct Answer: 5. (b, d)

Solution:

  1. 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, $\quad T _n>\frac{\pi}{3 \sqrt{3}}$



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