**Homework Problem on Series Convergence****Problem****a.** Prove \( \sum_{n=1}^{\infty} \frac{(-1)^n}{3n} \) converges conditionally, but not absolutely. Show all work.**b.** What value of \( n \) must you use to compute the \(n^{th}\) partial sum, \( S_n \), so that \( S_n \) is accurate to within 0.005? You do NOT need to calculate \( S_n \); just tell me what \( n \) works. Show all work.**Solution Steps**### Part a: Conditional Convergence Proof1. **Test for Absolute Convergence:** Consider the absolute value of the terms of the series: \[ \sum_{n=1}^{\infty} \left| \frac{(-1)^n}{3n} \right| = \sum_{n=1}^{\infty} \frac{1}{3n} \] This is a constant multiple of the harmonic series: \[ \sum_{n=1}^{\infty} \frac{1}{n} \] Since the harmonic series diverges, \(\sum_{n=1}^{\infty} \frac{1}{3n} \) also diverges. Therefore, the series does not converge absolutely.2. **Test for Conditional Convergence:** Use the Alternating Series Test (Leibniz's Test) for the series \( \sum_{n=1}^{\infty} \frac{(-1)^n}{3n} \). The Alternating Series Test requires checking two conditions: - The terms \( a_n = \frac{1}{3n} \) decrease monotonically. - \(\lim_{n \to \infty} a_n = 0 \) Both conditions are satisfied: - \( a_{n+1} = \frac{1}{3(n+1)} < \frac{1}{3n} = a_n \) for all \( n \), so the terms decrease monotonically. - \(\lim_{n \to \infty} \frac{1}{3n} = 0\) Thus, by the Alternating Series Test, the series \( \sum_{n=1

Name: **Homework Problem on Series Convergence****Problem****a.** Prove \(
Uploaded: 2024-03-20T06:38:58.000Z
Channel: Bartleby
Description: **Homework Problem on Series Convergence****Problem****a.** Prove \( \sum_{n=1}^{\infty} \frac{(-1)^n}{3n} \) converges conditionally, but not absolutely. Show all work.**b.** What value of \( n \) must you use to compute the \(n^{th}\) partial sum,