Question

For the series with \(n^{th}\) term \(x_n = \frac{n}{(n+1)(n+2)}\), the following is true: 

• A. The series \(\sum x_n\) is divergent
• B. The series \(\sum x_n\) is convergent
• C. The sequence of partial sums is convergent
• D. \(\lim_{n \to \infty} x_n = 0\)

Hint:

Exam Tip:

• If \(x_n \sim \frac{1}{n}\), the series diverges.
• The limit of the terms going to 0 is necessary but not sufficient for convergence.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to determine the convergence or divergence of the series \(\sum x_n\) where \(x_n = \frac{n}{(n+1)(n+2)}\).

Step 2: Key Formula or Approach:
Use the Limit Comparison Test with \(\frac{1}{n}\).

Step 3: Detailed Explanation:
\[ x_n = \frac{n}{n^2 + 3n + 2} \sim \frac{n}{n^2} = \frac{1}{n} \] We know that \(\sum \frac{1}{n}\) is divergent.
Using the limit comparison test: \[ \lim_{n \to \infty} \frac{x_n}{1/n} = \lim_{n \to \infty} \frac{n^2}{(n+1)(n+2)} = 1 \] Since the limit is a positive finite number, the series \(\sum x_n\) behaves like \(\sum \frac{1}{n}\), which is divergent.
So, the series is divergent.
Also, \(\lim_{n \to \infty} x_n = \lim_{n \to \infty} \frac{n}{n^2} = 0\).
But the series diverges even though the terms tend to 0.

Step 4: Final Answer:
Therefore, option (A) is correct.