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: Finding $a$:

$$a = \lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^{n-1} k = \lim_{n \to \infty} \frac{1}{n^2} \cdot \frac{(n-1)n}{2}$$

$$= \lim_{n \to \infty} \frac{n^2 - n}{2n^2} = \lim_{n \to \infty} \frac{1 - \frac{1}{n}}{2} = \frac{1}{2}$$

Step 2: Finding $b$:

$$b = \lim_{n \to \infty} \sum_{k=n+1}^{2n} \frac{1}{k} = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{n+k}$$

$$= \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{n} \cdot \frac{1}{1 + \frac{k}{n}}$$

This is a Riemann sum for $\int_0^1 \frac{1}{1+x} dx$ with partition width $\frac{1}{n}$:

$$b = \int_0^1 \frac{1}{1+x} dx = [\ln(1+x)]_0^1 = \ln 2 - \ln 1 = \ln 2$$

Step 3: Evaluating the options:

(A) $a > b$: $$\frac{1}{2} \stackrel{?}{>} \ln 2$$

Since $\ln 2 \approx 0.693 > 0.5$, this is FALSE

(B) $a < b$: $$\frac{1}{2} < \ln 2$$

This is TRUE

(C) $ab = \ln \sqrt{2}$: $$ab = \frac{1}{2} \cdot \ln 2 = \frac{\ln 2}{2} = \ln 2^{1/2} = \ln \sqrt{2}$$

This is TRUE

(D) $\frac{a}{b} = \ln \sqrt{2}$: $$\frac{a}{b} = \frac{1/2}{\ln 2} = \frac{1}{2\ln 2}$$

We need to check if $\frac{1}{2\ln 2} = \ln \sqrt{2} = \frac{\ln 2}{2}$

This would require $\frac{1}{2\ln 2} = \frac{\ln 2}{2}$, or $1 = (\ln 2)^2$

Since $(\ln 2)^2 \approx 0.48 \neq 1$, this is FALSE

Answer: (B) and (C) are true