Question

The necessary condition of a convergent sequence is :

• A. Bounded
• B. Having more than one limit
• C. Monotonic
• D. Not Cauchy

Hint:

Remember: - Convergent \(\implies\) Bounded (Necessary).
- Bounded \(\not\implies\) Convergent (e.g., \(\{a_n\} = (-1)^n\) is bounded but diverges).
Therefore, boundedness is a necessary but not sufficient condition for convergence.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question asks for a necessary condition for a sequence of real numbers to be convergent.
A necessary condition is one that must be true if the sequence converges, though it may not be sufficient on its own to guarantee convergence.
Key Formula or Approach: Recall the definitions and basic theorems of real analysis concerning convergent sequences:
- Theorem: Every convergent sequence of real numbers is bounded.

Step 2: Detailed Explanation:

• Let \(\{a_n\}\) be a convergent sequence in \(\mathbb{R}\) that converges to a limit \(L\).
By definition of convergence, for \(\varepsilon = 1\), there exists an integer \(N \in \mathbb{N}\) such that for all \(n \ge N\):
\[ |a_n - L| < 1 \] which implies:
\[ L - 1 < a_n < L + 1 \]

• Let \(M = \max\{a_1, a_2, \dots, a_{N-1}, L+1\}\) and \(m = \min\{a_1, a_2, \dots, a_{N-1}, L-1\}\).
Then, for all \(n \in \mathbb{N}\), we have:
\[ m \le a_n \le M \] This proves that the sequence \(\{a_n\}\) is bounded.

• Thus, being bounded is a necessary condition for a sequence to converge.

• Let us evaluate the other options: - Option (B) "Having more than one limit" is incorrect because a convergent sequence must have a unique limit.
- Option (C) "Monotonic" is incorrect because a sequence can converge without being monotonic (e.g., \(a_n = \frac{(-1)^n}{n}\) is oscillatory but converges to 0).
- Option (D) "Not Cauchy" is incorrect because every convergent sequence in \(\mathbb{R}\) is a Cauchy sequence (by completeness of \(\mathbb{R}\)).

Step 3: Final Answer:
The necessary condition of a convergent sequence is that it must be bounded.