SATHEE: Indefinite Integration 4 Question 5

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 $n = 1, 2, 3, \dots$ , then

(a) $S_{n} < \frac{π}{3 \sqrt{3}}$

(b) $S_{n} > \frac{π}{3 \sqrt{3}}$

(d) $T_{n} > \frac{π}{3 \sqrt{3}}$

$(2008, 4$ M)

Analytical & Descriptive Question

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Answer:

Correct Answer: 5. (b, d)

Solution:

Given, $S_{n} = \sum_{k = 0}^{n} \frac{n}{n^{2} + k n + k^{2}}$

$\begin{aligned} = \sum_{k = 0}^{n} \frac{1}{n} \cdot \frac{1}{1 + \frac{k}{n} + \frac{k^{2}}{n^{2}}} < lim_{n \to \infty} \sum_{k = 0}^{n} \frac{1}{n} \frac{1}{1 + \frac{k}{n} + \frac{k^{2}}{n}} \\ = \int_{0}^{1} \frac{1}{1 + x + x^{2}} d x \\ = \frac{2}{\sqrt{3}} \tan^{- 1} \frac{2}{\sqrt{3}} x + \frac{1}{2} \\ = \frac{2}{\sqrt{3}} \cdot \frac{π}{3} - \frac{π}{6} = \frac{π}{3 \sqrt{3}} i.e. S_{n} < \frac{π}{3 \sqrt{3}} \end{aligned}$

Similarly, $T_{n} > \frac{π}{3 \sqrt{3}}$

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 $n = 1, 2, 3, \dots$ , then

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Indefinite Integration 4 Question 5

5. Let Sn=∑k=0nnn2+kn+k2 and Tn=∑k=0n−1nn2+kn+k2, for n=1,2,3,…, then

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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 $n = 1, 2, 3, \dots$ , then