Question

$\lim_{n \to \infty} \left(\frac{n}{n^2 + 1^2} + \frac{n}{n^2 + 2^2} + \dots + \frac{n}{n^2 + 3^2 + \dots + \frac{1}{5n}} \right) =$

• A. $\frac{\pi}{4}$
• B. $\tan^{-1} 3$
• C. $\tan^{-1} 2$
• D. $\frac{\pi}{2}$

Answer 1

The Correct Option is C

1. Understand the limit:

We need to evaluate: \[ \lim_{n \to \infty} \left( \frac{n}{n^2 + 1^2} + \frac{n}{n^2 + 2^2} + \cdots + \frac{1}{5n} \right) \]

2. Recognize the sum as a Riemann sum:

The given expression can be written as: \[ \lim_{n \to \infty} \sum_{k=1}^{4n} \frac{n}{n^2 + k^2} \] This resembles the Riemann sum for an integral.

3. Convert to an integral:

Divide numerator and denominator by \( n^2 \): \[ \frac{n}{n^2 + k^2} = \frac{1/n}{1 + (k/n)^2} \] The sum becomes: \[ \lim_{n \to \infty} \sum_{k=1}^{4n} \frac{1}{n} \cdot \frac{1}{1 + (k/n)^2} \] This is the Riemann sum for: \[ \int_{0}^{4} \frac{1}{1 + x^2} dx \]

4. Evaluate the integral:

\[ \int_{0}^{4} \frac{1}{1 + x^2} dx = \tan^{-1} x \Big|_{0}^{4} = \tan^{-1} 4 - \tan^{-1} 0 = \tan^{-1} 4 \] However, the upper limit should be 2 (since \( k \) goes up to \( 2n \) when the last term is \( \frac{1}{5n} \)): \[ \int_{0}^{2} \frac{1}{1 + x^2} dx = \tan^{-1} 2 \]

Correct Answer: (C) \( \tan^{-1} 2 \)