Question

For \( n \in \mathbb{N} \), let \[ x_n = \sum_{k=1}^{n} \frac{k}{n^2 + k}. \] Then, which of the following is/are TRUE?

• A. The sequence \( (x_n) \) converges.
• B. The series \( \sum_{n=1}^{\infty} x_n \) converges.
• C. The series \( \sum_{n=1}^{\infty} x_n \) does NOT converge.
• D. The series \( \sum_{n=1}^{\infty} x_n^n \) converges.

Hint:

For a series to converge, its terms must approach zero as \( n \to \infty \). If the terms do not tend to zero, the series diverges.

Answer 1

The Correct Option is C

Step 1: Analyze the sequence \( x_n \).
We first express \( x_n \) as: \[ x_n = \sum_{k=1}^{n} \frac{k}{n^2 + k}. \] For large \( n \), the term \( \frac{k}{n^2 + k} \) behaves as: \[ \frac{k}{n^2 + k} \approx \frac{k}{n^2}. \] Thus, we approximate the sum: \[ x_n \approx \sum_{k=1}^{n} \frac{k}{n^2} = \frac{1}{n^2} \sum_{k=1}^{n} k = \frac{1}{n^2} \cdot \frac{n(n+1)}{2}. \] Simplifying, we get: \[ x_n \approx \frac{n(n+1)}{2n^2} = \frac{n+1}{2n}. \] As \( n \to \infty \), \( x_n \to \frac{1}{2} \). Step 2: Analyze the series \( \sum_{n=1}^{\infty} x_n \).
Since \( x_n \to \frac{1}{2} \) as \( n \to \infty \), the series \( \sum_{n=1}^{\infty} x_n \) does not converge, because the terms do not approach zero. Hence, the series diverges. Final Answer: \[ \boxed{\text{The series } \sum_{n=1}^{\infty} x_n \text{ does NOT converge.}} \]