Question:
For \(n\ \in \N\), let
\(a_n=\frac{1}{(3n+2)(3n+4)}\) and \(b_n=\frac{n^3+\cos(3^n)}{3n+n^3}\).
Then, which one of the following is TRUE ?
For \(n\ \in \N\), let
\(a_n=\frac{1}{(3n+2)(3n+4)}\) and \(b_n=\frac{n^3+\cos(3^n)}{3n+n^3}\).
Then, which one of the following is TRUE ?
\(a_n=\frac{1}{(3n+2)(3n+4)}\) and \(b_n=\frac{n^3+\cos(3^n)}{3n+n^3}\).
Then, which one of the following is TRUE ?
Updated On: Oct 22, 2024
- \(\sum\limits_{n=1}^{\infin}a_n\) is convergent but \(\sum\limits_{n=1}^{\infin}b_n\) is divergent
- \(\sum\limits_{n=1}^{\infin}a_n\) is divergent but \(\sum\limits_{n=1}^{\infin}b_n\) is convergent
- Both \(\sum\limits_{n=1}^{\infin}a_n\) and \(\sum\limits_{n=1}^{\infin}b_n\) are divergent
- Both \(\sum\limits_{n=1}^{\infin}a_n\) and \(\sum\limits_{n=1}^{\infin}b_n\) are convergent
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The Correct Option is D
Solution and Explanation
The correct option is (D) : Both \(\sum\limits_{n=1}^{\infin}a_n\) and \(\sum\limits_{n=1}^{\infin}b_n\) are convergent.
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