Question

Let \((a_n) = r^n\), where \(r\) be any real number, then

• A. \((a_n)\) is convergent for any \(r\)
• B. \((a_n)\) is divergent for any \(r\)
• C. \((a_n)\) is convergent for any \(|r| < 1\)
• D. \((a_n)\) is divergent for any \(|r| < 1\)

Hint:

For the geometric sequence \( r^n \):
- Converges to 0 if \( |r| < 1 \).
- Converges to 1 if \( r = 1 \).
- Diverges in all other cases.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This question concerns the convergence of a geometric sequence \( a_n = r^n \), where \( r \) is a real number.
The behavior of the term \( r^n \) as \( n \to \infty \) depends entirely on the value of the common ratio \( r \).

Step 2: Detailed Explanation:
Let us analyze the behavior of the limit \( \lim_{n \to \infty} r^n \) for all possible real values of \( r \):
- Case 1: \( |r| < 1 \) (which means \( -1 < r < 1 \))
In this case, the absolute value of the common ratio is less than 1. As we multiply a number less than 1 by itself repeatedly, it becomes progressively smaller.
Mathematically, we can prove this. Let \( |r| = \frac{1}{1 + h} \) where \( h > 0 \). Then:
\[ |r^n| = |r|^n = \frac{1}{(1 + h)^n} \] Using Bernoulli's inequality, we know that \( (1 + h)^n \ge 1 + nh \). Thus:
\[ 0 < |r^n| \le \frac{1}{1 + nh} \] Taking the limit as \( n \to \infty \):
\[ \lim_{n \to \infty} \frac{1}{1 + nh} = 0 \] By the Squeeze Theorem, \( \lim_{n \to \infty} r^n = 0 \).
Since the limit exists and is a finite real number (0), the sequence converges for all \( |r| < 1 \).
- Case 2: \( r = 1 \)
The sequence is \( (1, 1, 1, \dots) \), which trivially converges to 1.
- Case 3: \( r > 1 \)
The terms grow infinitely large, so \( \lim_{n \to \infty} r^n = \infty \), meaning the sequence diverges.
- Case 4: \( r \le -1 \)
If \( r = -1 \), the sequence oscillates between \( -1 \) and \( 1 \), which diverges. If \( r < -1 \), the sequence oscillates with infinitely growing amplitude, which also diverges.
Summarizing these cases, the sequence \( (a_n) = r^n \) converges if and only if \( -1 < r \le 1 \). Since \( |r| < 1 \) is a subset of this convergence interval, the sequence is always convergent for any \( |r| < 1 \).

Step 3: Final Answer:
The sequence \( (a_n) \) is convergent for any \( |r| < 1 \).
Therefore, the correct choice is Option (C).