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: Concept Overview:
A sequence's convergence is determined by evaluating the limit of its general term, \(a_n\), as \(n\) approaches infinity. A finite limit indicates convergence; an infinite or non-existent limit indicates divergence.

Step 2: Core Formula:
Evaluate the following limit: \[ \lim_{n \to \infty} a_n \]
Step 3: Step-by-Step Solution:
Given the sequence \(a_n = \frac{1}{n^2}\), we find the limit as \(n\) approaches infinity:\[ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{n^2} \]As \(n\) tends towards infinity, \(n^2\) also increases without bound, causing \(\frac{1}{n^2}\) to approach 0.\[ \lim_{n \to \infty} \frac{1}{n^2} = 0 \]
Step 4: Conclusion:
Because the sequence's limit is 0, a finite value, the sequence converges.