Question

Let \( a_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \) and \( b_n = \sum_{k=1}^{n} \frac{1}{k^2} \) for all \( n \in \mathbb{N} \). Then

• A. \( (a_n) \) is a Cauchy sequence but \( (b_n) \) is NOT a Cauchy sequence
• B. \( (a_n) \) is NOT a Cauchy sequence but \( (b_n) \) is a Cauchy sequence
• C. both \( (a_n) \) and \( (b_n) \) are Cauchy sequences
• D. neither \( (a_n) \) nor \( (b_n) \) is a Cauchy sequence

Answer 1

The Correct Option is B

1. For \( a_n \): - The sequence \( \{a_n\} \) represents the partial sums of the harmonic series. 
Although \( a_n \to \infty \) as \( n \to \infty \), the differences \( |a_{n+1} - a_n| = \frac{1}{n+1} \to 0 \). 
Hence, \( \{a_n\} \) is not a Cauchy sequence. 

2. For \( b_n \): - The sequence \( \{b_n\} = \frac{n^2}{2^n} \) tends to 0 as \( n \to \infty \). 
However, the differences \( |b_{n+1} - b_n| = \left|\frac{(n+1)^2}{2^{n+1}} - \frac{n^2}{2^n}\right| \) do not approach 0 because \( 2^{-n} \) dominates the numerator growth. 
Therefore, \( \{b_n\} \) is a Cauchy sequence.