Question

Let \((a_n)=n^2\) and \((b_n)=1+\frac{1}{n^2}\) be sequences of real numbers, then:

• A. Both ($\text{a}_\text{n}$) and ($\text{b}_\text{n}$) are convergent
• B. Both ($\text{a}_\text{n}$) and ($\text{b}_\text{n}$) are divergent
• C. ($\text{a}_\text{n}$) is convergent and ($\text{b}_\text{n}$) is divergent
• D. ($\text{a}_\text{n}$) is divergent and ($\text{b}_\text{n}$) is convergent

Hint:

- Any sequence that grows infinitely large (like \( n^k \) for \( k > 0 \)) is divergent.
- Any sequence where the variable \( n \) only appears in the denominator of a fraction with a constant numerator (like \( \frac{1}{n^k} \)) will converge to 0 as \( n \to \infty \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A sequence \( (x_n) \) of real numbers is convergent if it has a finite real limit as \( n \to \infty \). Otherwise, the sequence is divergent.

Step 2: Detailed Explanation:
Let us analyze the limits of both sequences individually as \( n \to \infty \):
1. Analyzing the sequence \( (a_n) = n^2 \):
We evaluate the limit of \( a_n \) as \( n \) approaches infinity:
\[ \lim_{n \to \infty} a_n = \lim_{n \to \infty} n^2 = \infty \]
Since the limit is not a finite real number (it grows without bound), the sequence \( (a_n) \) is divergent (specifically, it diverges to infinity).
2. Analyzing the sequence \( (b_n) = 1 + \frac{1}{n^2} \):
We evaluate the limit of \( b_n \) as \( n \) approaches infinity:
\[ \lim_{n \to \infty} b_n = \lim_{n \to \infty} \left( 1 + \frac{1}{n^2} \right) \]
Using the sum rule for limits:
\[ \lim_{n \to \infty} b_n = 1 + \lim_{n \to \infty} \frac{1}{n^2} \]
Since \( \lim_{n \to \infty} \frac{1}{n^2} = 0 \), we have:
\[ \lim_{n \to \infty} b_n = 1 + 0 = 1 \]
Since the limit is a finite real number (equal to 1), the sequence \( (b_n) \) is convergent.
Therefore, \( (a_n) \) is divergent and \( (b_n) \) is convergent.

Step 3: Final Answer:
The correct statement is ($\text{a}_\text{n}$) is divergent and ($\text{b}_\text{n}$) is convergent (Option D).