limn→∞[1n+1n+1+1n+2+⋯+12n]= &#x

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Question

{lim}_{n \to \infty} [\frac{1}{n} + \frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{2 n}] =

[Karnataka CET 1999]

Options:

A . 0

B . loge 4

C . loge 3

D . loge 2

Answer: Option D
:
D

{lim}_{n \to \infty} [\frac{1}{n} + \frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{2 n}]

= {lim}_{n \to \infty} [\frac{1}{n} + \frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{n + n}]

= \frac{1}{n} {lim}_{n \to \infty} [1 + \frac{1}{1 + \frac{1}{n}} + \frac{1}{1 + \frac{2}{n}} + \dots + \frac{1}{1 + \frac{n}{n}}]

= \frac{1}{n} {lim}_{n \to \infty} \sum_{r = 0}^{n} [\frac{1}{1 + \frac{r}{n}}] = \int_{0}^{1} \frac{1}{1 + x} d x

= [l o g_{e} (1 + x)]_{0}^{1} = l o g_{e} 2 - l o g_{e} 1 = l o g_{e} 2

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