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. $$ a_{n}=-\frac{2^{n}}{n}, n \geq 1 $$

Question

UpStudy AI · Accepted Answer

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. $$ a_{n}=\frac{1}{n}, n \geq 1 $$**.
The correct answer is C: $$ a_{n}=\frac{1}{n}, n \geq 1 $$.

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. \( a_{n}=-\frac{2^{n}}{n}, n \geq 1 \)

Upstudy AI Solution

Answer

Solution

Mind Expander

Related Questions

Latest Calculus Questions