Question

The sequence \(\{a_n = \frac{1}{n^2}; n>0\}\) is

• A. convergent
• B. divergent
• C. oscillates finitely
• D. oscillates infinitely

Hint:

For sequences of the form \(\frac{1}{n^p}\), if \(p>0\), the limit as \(n \to \infty\) is always 0, meaning the sequence converges to 0.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To determine if a sequence is convergent or divergent, we need to find the limit of the general term \(a_n\) as \(n\) approaches infinity. If the limit is a finite number, the sequence is convergent. If the limit is infinite or does not exist, the sequence is divergent.

Step 2: Key Formula or Approach:
We need to evaluate the limit: \[ \lim_{n \to \infty} a_n \]
Step 3: Detailed Explanation:
The given sequence is \(a_n = \frac{1}{n^2}\).
We calculate the limit as \(n \to \infty\): \[ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{n^2} \] As \(n\) becomes very large, \(n^2\) also becomes very large. Consequently, the fraction \(\frac{1}{n^2}\) approaches 0. \[ \lim_{n \to \infty} \frac{1}{n^2} = 0 \]
Step 4: Final Answer:
Since the limit of the sequence is 0, which is a finite value, the sequence is convergent.