lim n [ 1 1 + n + 1 2 + n + 1 3 + n + … + 1 2 n ] is equal to [2023]

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Solution

Q.
$\lim_{n \to \infty} [\frac{1}{1 + n} + \frac{1}{2 + n} + \frac{1}{3 + n} + \dots + \frac{1}{2 n}] is equal to$ [2023]

1 $\log_{e} 2$

2 $\log_{e} (\frac{2}{3})$

3 0

4 $\log_{e} (\frac{3}{2})$

Ans.
(1)

$\lim_{n \to \infty} [\frac{1}{1 + n} + \frac{1}{2 + n} + \frac{1}{3 + n} + \dots + \frac{1}{2 n}]$

$= \lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{r + n} = \lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} \frac{1}{1 + \frac{r}{n}} = \int_{0}^{1} \frac{d x}{1 + x} = \ln [1 + x] |_{0}^{1} = \ln 2$

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