Question

Let (an) be a sequence of real numbers defined by
\(a_n=\begin{cases} 1 & \text{if } n \text{ is prime}\\    -1 & \text{if } n \text{ is not prime} \end{cases}\)
Let \(b_n=\frac{a_n}{n}\) for n ∈ \(\N\). Then

• A. both (a_n) and (b_n) are convergent
• B. (a_n) is convergent but (b_n) is NOT convergent
• C. (a_n) is NOT convergent but (b_n) is convergent
• D. both (a_n) and (b_n) are NOT convergent

Answer 1

The Correct Option is C

To determine the convergence of the sequences \((a_n)\) and \((b_n)\), we need to analyze them individually.

1. Analysis of the sequence \((a_n)\):
   • The sequence \((a_n)\) is defined as: \(a_n = \begin{cases} 1 & \text{if } n \text{ is prime} \\ -1 & \text{if } n \text{ is not prime} \end{cases}\)
   • This sequence alternates between 1 and -1 depending on whether \(n\) is prime or not. Since prime numbers are not regularly spaced, there is no consistent pattern in the sequence values.
   • Because \((a_n)\) does not settle to a single value (since it does not settle to a limit, as prime distribution is unpredictable), it cannot be convergent.
2. Analysis of the sequence \((b_n)\):
   • The sequence \((b_n)\) is defined as: \(b_n = \frac{a_n}{n}\)
   • We analyze the limit of \((b_n)\) as \(n\) approaches infinity.
   • For any \(n\), \(\left|b_n\right| = \left|\frac{a_n}{n}\right| = \frac{1}{n}\). As \(n\) increases, \(\frac{1}{n}\) approaches 0.
   • Thus, \(\lim_{n \to \infty} b_n = 0\).
   • Since \((b_n)\) approaches a finite limit (0), it is convergent.

Therefore, the correct conclusion is that \((a_n)\) is not convergent, but \((b_n)\) is convergent.