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 ?

• A. \(\sum\limits_{n=1}^{\infin}a_n\) is convergent but \(\sum\limits_{n=1}^{\infin}b_n\) is divergent
• B. \(\sum\limits_{n=1}^{\infin}a_n\) is divergent but \(\sum\limits_{n=1}^{\infin}b_n\) is convergent
• C. Both \(\sum\limits_{n=1}^{\infin}a_n\) and \(\sum\limits_{n=1}^{\infin}b_n\) are divergent
• D. Both \(\sum\limits_{n=1}^{\infin}a_n\) and \(\sum\limits_{n=1}^{\infin}b_n\) are convergent

Answer 1

The Correct Option is D

- For \( a_n \): The term \( a_n \) is given by: \[ a_n = \frac{1}{(3n + 2)(3n + 4)}. \] As \( n \) becomes large, the dominant terms in the denominator are \( 3n \), so for large \( n \), \[ a_n \approx \frac{1}{9n^2}. \] The series \( \sum_{n=1}^{\infty} a_n \) behaves like the series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), which is a convergent p-series with \( p = 2 \). Hence, the series \( \sum_{n=1}^{\infty} a_n \) is convergent. - For \( b_n \): The term \( b_n \) is given by: \[ b_n = \frac{n^3 + \cos(3^n)}{3n^3 + n^3} = \frac{n^3 + \cos(3^n)}{4n^3}. \] For large \( n \), the \( \cos(3^n) \) term oscillates between -1 and 1, but it is dominated by the \( n^3 \) term in the numerator. So, for large \( n \), \[ b_n \approx \frac{n^3}{4n^3} = \frac{1}{4}. \] Thus, the series \( \sum_{n=1}^{\infty} b_n \) behaves like a constant series, which converges. Therefore, the series \( \sum_{n=1}^{\infty} b_n \) is convergent. Thus, both \( \sum_{n=1}^{\infty} a_n \) and \( \sum_{n=1}^{\infty} b_n \) are convergent. The correct answer is (D).