Question

Consider the following infinite series: \[ S_1 := \sum_{n=0}^{\infty} (-1)^n \frac{n}{n^2 + 4} \quad \text{and} \quad S_2 := \sum_{n=0}^{\infty} (-1)^n \sqrt{n^2 + 1 - n}. \] Which of the above series is/are conditionally convergent?

• A. \( S_1 \) only
• B. \( S_2 \) only
• C. Both \( S_1 \) and \( S_2 \)
• D. Neither \( S_1 \) nor \( S_2 \)

Hint:

To test if an alternating series is conditionally convergent, check if the terms tend to zero and if the sequence is monotonically decreasing.

Answer 1

The Correct Option is C

We need to determine whether the two given series are conditionally convergent.
Step 1: Analyze \( S_1 \).
The series \( S_1 \) is of the form: \[ S_1 = \sum_{n=0}^{\infty} (-1)^n \frac{n}{n^2 + 4} \] We can apply the alternating series test (Leibniz's test) to determine if this series is conditionally convergent. The alternating series test requires two conditions: 1. The terms \( \frac{n}{n^2 + 4} \) must decrease monotonically. 2. The limit of the terms must be zero as \( n \to \infty \). Let's check these: - The sequence \( \frac{n}{n^2 + 4} \) clearly tends to zero as \( n \to \infty \), because: \[ \lim_{n \to \infty} \frac{n}{n^2 + 4} = 0. \] - To check for monotonicity, consider the derivative of \( \frac{n}{n^2 + 4} \). The derivative is negative for \( n \geq 1 \), meaning the terms are decreasing for large \( n \). Since both conditions of the alternating series test are satisfied, \( S_1 \) converges conditionally.
Step 2: Analyze \( S_2 \).
The series \( S_2 \) is: \[ S_2 = \sum_{n=0}^{\infty} (-1)^n \sqrt{n^2 + 1 - n} \] We can again apply the alternating series test. First, observe that as \( n \to \infty \): \[ \lim_{n \to \infty} \sqrt{n^2 + 1 - n} \approx \lim_{n \to \infty} \sqrt{n^2} = n, \] which does not tend to zero. Hence, the limit of the terms is not zero, and the series does not converge. Therefore, \( S_2 \) does not converge conditionally or absolutely.