Q18 / 116 · NTA Abhyas (JEE Main) 2020 · Mathematics · Definite Integration

The value of $Ι = \underset{n \to \infty}{l i m} \sum_{r = n}^{2 n} \frac{n}{r^{2}}$ is equal to

A
$\frac{1}{3}$
Correct answer
B
$\frac{1}{2}$
C
$\frac{π}{2}$
D
$\frac{π}{3}$

Solution

Ι = \underset{n \to \infty}{l i m} \sum_{r = n}^{2 n} \frac{1}{\frac{r^{2}}{n^{2}}} \cdot \frac{1}{n}

= \int_{1}^{2} \frac{1}{x^{2}} d x = {(- \frac{1}{x})}_{1}^{2}

= (- \frac{1}{2}) - (- 1)

= \frac{1}{2}

Hence, the correct answer is option: 'B'.

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