Limits
Limit of a Sum
JEE Advanced 2013
Grade 11

Question:

$\lim_{n \to \infty} \left(\frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{3n}\right) = \ln k$. Find the value of $k$.

Step-by-Step Solution

Key Concept: Write the sum as $\frac{1}{n} \sum_{r=1}^{2n} \frac{1}{1 + r/n}$. This is a Riemann sum for $\int_0^2 \frac{dx}{1 + x} = \ln 3$. So $k = 3$.
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Correct Answer: 3

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