91Ó°ÊÓ

To check the convergence of this series, apply the alternating series test. As \(\frac{1}{k}\) is positive, decreasing for k=1,2... and approaches 0 as \(k\to\infty\), the sine function alternates between positive and negative values. So the condition for the alternating series test is met: 1. \(|\sin \frac{1}{k}|\) is decreasing. 2. \({\lim}_{k\to\infty} |\sin \frac{1}{k}| = 0\). Therefore, the series \(\sum_{k=1}^{\infty} \sin \frac{1}{k}\) converges under the alternating series test.

Now, let's analyze the second series. We can use the comparison test based on absolute convergence. Since, \(|\sin x| \le 1\) for every x, we have: \(0\le|\frac{1}{k} \sin \frac{1}{k}|\le\frac{1}{k}\) Now, we know that the series \(\sum_{k=1}^{\infty} \frac{1}{k}\) is a harmonic series and it diverges. So, you can write: \(\sum_{k=1}^{\infty}\frac{1}{k}\) diverges. Since \(0\le|\frac{1}{k} \sin \frac{1}{k}|\le\frac{1}{k}\) and \(\sum_{k=1}^{\infty}\frac{1}{k}\) diverges, by the comparison test, the series \(\sum_{k=1}^{\infty}\frac{1}{k} \sin \frac{1}{k}\) also diverges. To summarize, we have determined the convergence of the two given series: a. \(\sum_{k=1}^{\infty} \sin \frac{1}{k}\) converges. b. \(\sum_{k=1}^{\infty} \frac{1}{k} \sin \frac{1}{k}\) diverges.