The value of l i m n → ∞ 1 2 n + 1 2 n + 1 + 1 2 n + 2 + . . . . . + 1 4 n is equal to

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Question: Answered & Verified by Expert

The value of

\underset{n \to \infty}{l i m} (\frac{1}{2 n} + \frac{1}{2 n + 1} + \frac{1}{2 n + 2} + . . . . . + \frac{1}{4 n})

is equal to

MathematicsDefinite IntegrationJEE Main

Options:

A $e^{2}$
B $\ln 2$
C $\ln 4$
D $3 \ln 2$

Solution:

1482 Upvotes Verified Answer

The correct answer is:

\ln 2

\underset{n \to \infty}{l i m} \sum_{r = 0}^{2 n} \frac{1}{2 n + r} = \underset{n \to \infty}{l i m} \frac{1}{n} \sum_{r = 0}^{2 n} \frac{1}{2 + \frac{r}{n}}

= \int_{0}^{2} (\frac{1}{2 + x}) d x = {|\ln (x + 2)|}_{0}^{2}

= \ln 4 - \ln 2 = \ln 2

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