Question

The value of \( \lim_{n\to\infty} \left[ \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{6n} \right] \) is

• A. \( \ln 3 \)
• B. \( \ln 6 \)
• C. \( e^3 \)
• D. \( e^6 \)
• E. \( \ln 2 \)

Hint:

Harmonic sums over proportional limits always give logarithms.

Answer 1

The Correct Option is B

Concept: A sum of the form \( \sum \frac{1}{k} \) can be approximated by a definite integral: \[ \sum_{k=a}^{b} \frac{1}{k} \approx \int_a^b \frac{dx}{x} \]

Step 1: Write the sum clearly. \[ \sum_{k=n+1}^{6n} \frac{1}{k} \]

Step 2: Convert into integral form. \[ \sum_{k=n+1}^{6n} \frac{1}{k} \approx \int_n^{6n} \frac{dx}{x} \]

Step 3: Evaluate the integral. \[ \int_n^{6n} \frac{dx}{x} = \ln(6n) - \ln(n) \] \[ = \ln\left(\frac{6n}{n}\right) \] \[ = \ln 6 \]

Step 4: Take the limit. As \(n \to \infty\), approximation becomes exact: \[ \lim_{n\to\infty} \sum_{k=n+1}^{6n} \frac{1}{k} = \ln 6 \]