SATHEE: Indefinite Integration 4 Question 2

Indefinite Integration 4 Question 2

2. $lim_{n \to \infty} \frac{n}{n^{2} + 1^{2}} + \frac{n}{n^{2} + 2^{2}} + \frac{n}{n^{2} + 3^{2}} + \dots + \frac{1}{5 n}$ is equal to

(a) $\tan^{- 1} (3)$

(b) $\tan^{- 1} (2)$

(d) $π / 2$

(2019 Main, 12 Jan II)

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Answer:

Correct Answer: 2. $(a, d)$

Solution:

Clearly,

$\begin{aligned} lim_{n \to \infty} \frac{n}{n^{2} + 1^{2}} + \frac{n}{n^{2} + 2^{2}} + \frac{n}{n^{2} + 3^{2}} + \dots + \frac{1}{5 n} \\ = lim_{n \to \infty} \frac{n}{n^{2} + 1^{2}} + \frac{n}{n^{2} + 2^{2}} + \frac{n}{n^{2} + 3^{2}} + \dots + \frac{n}{n^{2} + (2 n)^{2}} \\ = lim_{n \to \infty} \sum_{r = 1}^{2 n} \frac{n}{n^{2} + r^{2}} \end{aligned}$

$\begin{aligned} = lim_{n \to \infty} \sum_{r = 1}^{2 n} \frac{1}{1 + \frac{r}{n}^{2}} \cdot \frac{1}{n} = \int_{0}^{2} \frac{d x}{1 + x^{2}} \\ ∵ lim_{n \to \infty} \sum_{r = 1}^{p n} \frac{1}{n} f \frac{r}{n} = \int_{0}^{p} f (x) d x \\ = {[\tan^{- 1} x]}_{0}^{2} = \tan^{- 1} 2 \end{aligned}$

Indefinite Integration 4 Question 2

2. $lim_{n \to \infty} \frac{n}{n^{2} + 1^{2}} + \frac{n}{n^{2} + 2^{2}} + \frac{n}{n^{2} + 3^{2}} + \dots + \frac{1}{5 n}$ is equal to

Answer:

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Indefinite Integration 4 Question 2

2. limn→∞nn2+12+nn2+22+nn2+32+…+15n is equal to

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2. $lim_{n \to \infty} \frac{n}{n^{2} + 1^{2}} + \frac{n}{n^{2} + 2^{2}} + \frac{n}{n^{2} + 3^{2}} + \dots + \frac{1}{5 n}$ is equal to