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 \( \{ s_n \}_{n=1}^{\infty} \)? 
 

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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.
Updated On: Dec 15, 2025
  • \( \{ s_n \}_{n=1}^{\infty} \) converges in \( \mathbb{Q} \) 
     

  • \( \{ s_n \}_{n=1}^{\infty} \) is a Cauchy sequence but does not converge in \( \mathbb{Q} \) 
     

  • The subsequence \( \{ s_{k^n} \}^\infty_{n=1} \) is convergent in \( \mathbb{R} \), only when \( k \) is even natural number 
     

  • \( \{ s_n \}_{n=1}^{\infty} \) is not a Cauchy sequence 
     

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The Correct Option is B

Solution and Explanation

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) \( \{ s_n \}_{n=1}^{\infty}\) converges in \( \mathbb{Q} \): This is incorrect, since the limit of the sequence is \( e \), which is not a rational number. 
(B) \( \{ s_n \}_{n=1}^{\infty}\) 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 \( \{ s_{k^n} \}^\infty_{n=1} \) 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) \( \{ s_n \}_{n=1}^{\infty}\) is not a Cauchy sequence: This is incorrect, as \( \{ s_n \}_{n=1}^{\infty} \) is a Cauchy sequence in \( \mathbb{R} \). 
 

Step 3: Conclusion. 
The correct answer is (B) \( \{ s_n \}_{n=1}^{\infty}\) is a Cauchy sequence but does not converge in \( \mathbb{Q} \)
 

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