Question

Let \( u_n = \frac{(4 - n)}{n} \), \( n \in \mathbb{N} \), and let \( l = \lim_{n \to \infty} u_n \).
Which of the following statements is TRUE?

• A. \( l = 0 \) and \( \sum_{n=1}^{\infty} u_n \) is convergent
• B. \( l = 1 \) and \( \sum_{n=1}^{\infty} u_n \) is divergent
• C. \( l = \frac{1}{4} \) and \( \sum_{n=1}^{\infty} u_n \) is oscillatory
• D. \( l = \frac{1}{4} \) and \( \sum_{n=1}^{\infty} u_n \) is divergent

Hint:

For convergence of a series, the terms must approach zero as \( n \to \infty \). If the terms do not approach zero, the series will diverge.

Answer 1

The Correct Option is B

Step 1: Find the limit of the sequence \( u_n \).
We are given the sequence: \[ u_n = \frac{4 - n}{n} \] As \( n \to \infty \), we calculate the limit: \[ \lim_{n \to \infty} u_n = \lim_{n \to \infty} \frac{4 - n}{n} = \lim_{n \to \infty} \left( \frac{4}{n} - 1 \right) \] Since \( \frac{4}{n} \to 0 \) as \( n \to \infty \), we have: \[ \lim_{n \to \infty} u_n = 0 - 1 = -1 \] Thus, the limit of the sequence is \( l = -1 \).

Step 2: Analyze the series \( \sum_{n=1}^{\infty} u_n \).
We check the behavior of the series \( \sum_{n=1}^{\infty} u_n \). For large \( n \), we can approximate \( u_n \) as: \[ u_n \approx \frac{-n}{n} = -1 \] Since \( u_n \) approaches a nonzero constant, the series does not converge to a finite value. In fact, the series diverges.

Step 3: Final Answer.
The correct statement is:
\[ l = -1 \quad \text{and} \quad \sum_{n=1}^{\infty} u_n \text{ is divergent}. \]

Final Answer: \( l = -1 \) and \( \sum_{n=1}^{\infty} u_n \) is divergent.