Question

Let \((a_n) = (-1)^n\) and \((b_n) = -2^n\) be sequences of real numbers, then

• A. \((a_n)\) is convergent and \((b_n)\) is divergent
• B. \((a_n)\) is divergent and \((b_n)\) is convergent
• C. Both \((a_n)\) and \((b_n)\) are convergent
• D. Both \((a_n)\) and \((b_n)\) are divergent

Hint:

A sequence is only convergent if it approaches a single, finite real number as \( n \to \infty \).
If a sequence bounces back and forth between different values, or if it grows larger and larger without limit, it is always divergent.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This question deals with real analysis and the convergence properties of real-valued sequences.
A sequence of real numbers \( (x_n) \) is said to converge to a real number \( L \) if, for every \( \epsilon > 0 \), there exists an integer \( N \) such that for all \( n \ge N \), we have \( |x_n - L| < \epsilon \). If no such finite real number \( L \) exists, the sequence is classified as divergent.

Step 2: Detailed Explanation:
Let us analyze each of the given sequences individually:
- Sequence 1: \( (a_n) = (-1)^n \)
Let us write down the first few terms of this sequence:
\[ a_1 = -1, \quad a_2 = 1, \quad a_3 = -1, \quad a_4 = 1, \quad \dots \] This is an oscillating sequence that alternates between \( -1 \) and \( 1 \).
To test its convergence, let us look at its subsequences:
- The subsequence of even terms is \( a_{2k} = (-1)^{2k} = 1 \), which converges to \( 1 \) as \( k \to \infty \).
- The subsequence of odd terms is \( a_{2k-1} = (-1)^{2k-1} = -1 \), which converges to \( -1 \) as \( k \to \infty \).
According to real analysis, a sequence converges if and only if all of its subsequences converge to the same unique limit.
Since the two subsequences of \( (a_n) \) converge to two different limits (\( 1 \) and \( -1 \)), the parent sequence \( (a_n) \) does not have a unique limit.
Therefore, the sequence \( (a_n) \) is divergent (specifically, it oscillates finitely).
- Sequence 2: \( (b_n) = -2^n \)
Let us write down the first few terms of this sequence:
\[ b_1 = -2, \quad b_2 = -4, \quad b_3 = -8, \quad b_4 = -16, \quad \dots \] As \( n \) increases, the terms of the sequence decrease without bound.
Let us compute the limit as \( n \to \infty \):
\[ \lim_{n \to \infty} b_n = \lim_{n \to \infty} (-2^n) = -\infty \] Since the limit is not a finite real number, the sequence \( (b_n) \) diverges to negative infinity.
Therefore, the sequence \( (b_n) \) is divergent.
Both sequences are divergent.

Step 3: Final Answer:
Both \( (a_n) \) and \( (b_n) \) are divergent.
Therefore, the correct choice is Option (D).