Give an example of a nonincreasing sequence with a limit. Choose the correct answer below. A. \( a_{n}=(-1)^{n} n, n \geq 1 \) B. \( a_{n}=\frac{\sin n}{n}, n \geq 1 \) C. \( a_{n}=\frac{1}{n}, n \geq 1 \) D. \( a_{n}=-\frac{2^{n}}{n}, n \geq 1 \)

To determine which of the given sequences is a nonincreasing sequence with a limit, let's analyze each option step by step. 1. **Option A: \( a_{n}=(-1)^{n} n, n \geq 1 \)** - This sequence alternates between positive and negative values: \( -1, 2, -3, 4, -5, \ldots \). - It does not converge to a limit, as it oscillates indefinitely. 2. **Option B: \( a_{n}=\frac{\sin n}{n}, n \geq 1 \)** - The sine function oscillates between -1 and 1, so \( \frac{\sin n}{n} \) oscillates between \( -\frac{1}{n} \) and \( \frac{1}{n} \). - As \( n \) increases, \( \frac{\sin n}{n} \) approaches 0, but it is not nonincreasing because it oscillates. 3. **Option C: \( a_{n}=\frac{1}{n}, n \geq 1 \)** - This sequence is \( 1, \frac{1}{2}, \frac{1}{3}, \ldots \). - It is nonincreasing (each term is less than or equal to the previous term) and converges to 0 as \( n \) approaches infinity. 4. **Option D: \( a_{n}=-\frac{2^{n}}{n}, n \geq 1 \)** - This sequence is negative and grows in magnitude: \( -\frac{2}{1}, -\frac{4}{2}, -\frac{8}{3}, \ldots \). - It is nonincreasing, but it diverges to negative infinity, so it does not have a limit. Based on this analysis, the only sequence that is nonincreasing and has a limit is: **C. \( a_{n}=\frac{1}{n}, n \geq 1 \)**.