Question

Let \(a_n=\frac{1}{n^2}\) be a sequence of real numbers. Then the sequence is:

• A. Convergent and limit is 1
• B. Convergent and limit is 0
• C. Divergent
• D. Convergent and limits are 0 and 1

Hint:

For any positive real number \( p > 0 \):
\[ \lim_{n \to \infty} \frac{1}{n^p} = 0 \]
Thus, all such sequences are convergent and converge to 0.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We determine the convergence and limit of a basic rational sequence.

Step 2: Detailed Explanation:
Let the sequence be defined as:
\[ a_n = \frac{1}{n^2} \]
We evaluate the limit of the sequence as \( n \) approaches infinity:
\[ L = \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{n^2} \]
As \( n \) becomes extremely large, \( n^2 \) also grows infinitely large.
The reciprocal of an infinitely large positive number approaches zero:
\[ L = 0 \]
Since the limit \( L = 0 \) is a unique, finite real number, the sequence is convergent, and its limit is 0.
Note that according to the uniqueness of limits theorem, a convergent sequence can only have a single, unique limit, which rules out Option D.

Step 3: Final Answer:
The sequence is Convergent and limit is 0 (Option B).