Question

Let \( a = \lim_{n \to \infty} \left( \frac{1}{n^2} + \frac{2}{n^2} + \cdots + \frac{n-1}{n^2} \right) \) and \( b = \lim_{n \to \infty} \left( \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{n+n} \right). \) Which of the following is/are true?

• A. \( a > b \)
• B. \( a < b \)
• C. \( ab = \ln \sqrt{2} \)
• D. \( \dfrac{a}{b} = \ln \sqrt{2} \)

Hint:

Riemann sums involving reciprocals often converge to logarithmic integrals like \( \ln(2) \).

Answer 1

The Correct Option is B, C

Step 1: Compute \( a \).
\[ a = \lim_{n \to \infty} \frac{1}{n^2} (1 + 2 + 3 + \cdots + (n-1)) = \lim_{n \to \infty} \frac{1}{n^2} \cdot \frac{n(n-1)}{2} = \frac{1}{2}. \]

Step 2: Compute \( b \).
\[ b = \lim_{n \to \infty} \left( \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \right). \] This is a Riemann sum approximation of the integral \[ b = \int_1^2 \frac{1}{x} \, dx = \ln 2. \]

Step 3: Compare values.
\[ a = \frac{1}{2} = 0.5, b = \ln 2 \approx 0.693. \] Hence \( a < b. \)

Final Answer: \[ \boxed{a < b.} \]