Question:
Let \(a_n=\frac{1+2^{-2}+...+n^{-2}}{n}\) for n ∈ \(\N\). Then
Let \(a_n=\frac{1+2^{-2}+...+n^{-2}}{n}\) for n ∈ \(\N\). Then
Updated On: Oct 1, 2024
- both the sequence (an) and the series \(\sum\limits_{n=1}^{\infin}a_n\) are convergent
- the sequence (an) is convergent but the series \(\sum\limits_{n=1}^{\infin}a_n\) are NOT convergent
- both the sequence (an) and the series \(\sum\limits_{n=1}^{\infin}a_n\) are NOT convergent
- the sequence (an) is NOT convergent but the series \(\sum\limits_{n=1}^{\infin}a_n\) is convergent
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The Correct Option is B
Solution and Explanation
The correct option is (B) : the sequence (an) is convergent but the series \(\sum\limits_{n=1}^{\infin}a_n\) are NOT convergent.
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