Question

Let $(a_n) = (\frac{n+1}{n})$ and $(b_n) = (-1)^n$ be sequences of real number, then

• A. Both $(a_n)$ and $(b_n)$ are convergent
• B. Both $(a_n)$ and $(b_n)$ are divergent
• C. $(a_n)$ is convergent but $(b_n)$ is not convergent
• D. $(a_n)$ is not convergent but $(b_n)$ is convergent

Hint:

To test sequence convergence quickly in exams:
- If the terms approach a single constant value as \(n \to \infty\), it converges.
- If the terms oscillate between different values or grow to \(\pm\infty\), it diverges.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In real analysis, a sequence of real numbers \((x_n)\) is said to be convergent if its limit as \(n\) approaches infinity exists and is a unique real number \(L\):
\[ \lim_{n\to\infty} x_n = L \] If no such unique real limit exists, the sequence is divergent (not convergent).

Step 3: Detailed Explanation:
Let us analyze the convergence of both given sequences independently:
1. Analyzing the sequence \((a_n) = \left(\frac{n+1}{n}\right)\):
We can rewrite the general term as:
\[ a_n = \frac{n}{n} + \frac{1}{n} = 1 + \frac{1}{n} \] Now, let us find the limit as \(n \to \infty\):
\[ \lim_{n\to\infty} a_n = \lim_{n\to\infty} \left(1 + \frac{1}{n}\right) = 1 + 0 = 1 \] Because the limit is a unique real number (\(L = 1\)), the sequence \((a_n)\) is convergent.
2. Analyzing the sequence \((b_n) = (-1)^n\):
Let us write out the terms of this sequence:
\[ b_1 = -1,\ b_2 = 1,\ b_3 = -1,\ b_4 = 1, \dots \] The sequence oscillates infinitely between \(-1\) and \(1\).
The even terms converge to \(+1\), while the odd terms converge to \(-1\).
Because there are two distinct subsequential limits, the overall limit as \(n \to \infty\) does not exist.
Therefore, the sequence \((b_n)\) is divergent (not convergent).
Thus, \((a_n)\) is convergent, but \((b_n)\) is not convergent.

Step 4: Final Answer:
The correct statement is that \((a_n)\) is convergent but \((b_n)\) is not convergent, matching Option (C).