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
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
\(a_n=\begin{cases} 1 & \text{if } n \text{ is prime}\\ -1 & \text{if } n \text{ is not prime} \end{cases}\)
Let \(b_n=\frac{a_n}{n}\) for n ∈ \(\N\). Then
Updated On: Oct 1, 2024
- both (an) and (bn) are convergent
- (an) is convergent but (bn) is NOT convergent
- (an) is NOT convergent but (bn) is convergent
- both (an) and (bn) are NOT convergent
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The Correct Option is C
Solution and Explanation
The correct option is (C) : (an) is NOT convergent but (bn) is convergent.
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