Question

Let \(a_n=\sin(\frac{1}{n^3})\) and \(b_n=\sin(\frac{1}{n})\) for n ∈ \(\N\). Then

• A. both \(\sum\limits_{n=1}^{\infin}a_n\) and \(\sum\limits_{n=1}^{\infin}b_n\) are convergent
• B. \(\sum\limits_{n=1}^{\infin}a_n\) is convergent \(\sum\limits_{n=1}^{\infin}b_n\) is NOT convergent
• C. \(\sum\limits_{n=1}^{\infin}a_n\) is NOT convergent \(\sum\limits_{n=1}^{\infin}b_n\) is convergent
• D. both \(\sum\limits_{n=1}^{\infin}a_n\) and \(\sum\limits_{n=1}^{\infin}b_n\) are NOT convergent

Answer 1

The Correct Option is B

To determine the convergence of the series \(\sum\limits_{n=1}^{\infty}a_n\) and \(\sum\limits_{n=1}^{\infty}b_n\), where \(a_n = \sin\left(\frac{1}{n^3}\right)\) and \(b_n = \sin\left(\frac{1}{n}\right)\), we need to analyze their behaviors as \(n \rightarrow \infty\).

1. Convergence of \(\sum\limits_{n=1}^{\infty}a_n\): 

For large \(n\), \(\frac{1}{n^3}\) becomes very small. Using the approximation \(\sin(x) \approx x\) when \(x\) is near zero, we can approximate:

\(\sin\left(\frac{1}{n^3}\right) \approx \frac{1}{n^3}\)

Thus, \(a_n \approx \frac{1}{n^3}\)

The series \(\sum\limits_{n=1}^{\infty} \frac{1}{n^3}\) is a p-series with \(p = 3\), which is known to converge because \(p > 1\).

Conclusion: \(\sum\limits_{n=1}^{\infty}a_n\) converges.

2. Convergence of \(\sum\limits_{n=1}^{\infty}b_n\):

Similarly, for large \(n\), \(\frac{1}{n}\) is small. Using the approximation, we have:

\(\sin\left(\frac{1}{n}\right) \approx \frac{1}{n}\)

Thus, \(b_n \approx \frac{1}{n}\)

The series \(\sum\limits_{n=1}^{\infty} \frac{1}{n}\) is the harmonic series, which is known to diverge.

Conclusion: \(\sum\limits_{n=1}^{\infty}b_n\) diverges.

Final Answer: \(\sum\limits_{n=1}^{\infty}a_n\) is convergent, and \(\sum\limits_{n=1}^{\infty}b_n\) is NOT convergent. Thus, the correct option is that \(\sum\limits_{n=1}^{\infty}a_n\) is convergent, and \(\sum\limits_{n=1}^{\infty}b_n\) is NOT convergent.