Question

Let \( s_n = 1 + \frac{1}{1!} + \frac{1}{2!} + \cdots + \frac{1}{n!} \) for \( n \in \mathbb{N} \). Then which one of the following is TRUE for the sequence \( \{ s_n \} \)?

• A. \( \{ s_n \} \) converges in \( \mathbb{Q} \)
• B. \( \{ s_n \} \) is a Cauchy sequence but does not converge in \( \mathbb{Q} \)
• C. The subsequence \( \{ s_k \}^\infty_{k=1} \) is convergent in \( \mathbb{R} \), only when \( k \) is even natural number
• D. \( \{ s_n \} \) is not a Cauchy sequence

Hint:

A Cauchy sequence in \( \mathbb{R} \) may converge to an irrational number. However, it may not converge in \( \mathbb{Q} \), since rational numbers are not complete.

Answer 1

The Correct Option is B

Step 1: Understanding the sequence.
The sequence \( s_n = 1 + \frac{1}{1!} + \frac{1}{2!} + \cdots + \frac{1}{n!} \) is the partial sum of the series for the exponential function \( e^1 \), and it converges to \( e \), a transcendental number.

Step 2: Analyzing the options.
(A) \( \{ s_n \} \) converges in \( \mathbb{Q} \): This is incorrect, since the limit of the sequence is \( e \), which is not a rational number.
(B) \( \{ s_n \} \) is a Cauchy sequence but does not converge in \( \mathbb{Q} \): Correct — The sequence is Cauchy because it converges in \( \mathbb{R} \), but it does not converge to a rational number.
(C) The subsequence \( \{ s_k \}^\infty_{k=1} \) is convergent in \( \mathbb{R} \), only when \( k \) is even natural number: This is incorrect, as the entire sequence converges to \( e \), not just the subsequence.
(D) \( \{ s_n \} \) is not a Cauchy sequence: This is incorrect, as \( \{ s_n \} \) is a Cauchy sequence in \( \mathbb{R} \).

Step 3: Conclusion.
The correct answer is (B) \( \{ s_n \} \) is a Cauchy sequence but does not converge in \( \mathbb{Q} \).