Question

Use the Limit Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \sin \frac{1}{n} $$

Short answer

The given series \( \sum_{n=1}^{\infty} \sin \frac{1}{n} \) diverges.

Step-by-step solution

Identify a Comparing Series

Start by identifying a series that you can compare with the given series. The series \( \sum_{n=1}^{\infty} \sin \frac{1}{n} \) can be compared with the series \( \sum_{n=1}^{\infty} \frac{1}{n} \), where 1/n is the argument of the sine function.

Apply Limit Comparison Test

Apply the limit comparison test which involves taking the limit as n approaches infinity of the ratio of the term of the given series to the corresponding term of the comparison series. So, we consider \[\lim_{n \to \infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}}\]In order to simplify this expression, we can use the limit identity \(\lim_{n \to \infty} \frac{\sin x}{x} = 1\), which leaves us with 1.

Drawing Conclusions

According to the limit comparison test, if the limit of the ratio of corresponding terms of two positive series (as n approaches infinity) is a finite nonzero number, then either both series converge or both diverge. Here, the comparison series, \( \sum_{n=1}^{\infty} \frac{1}{n} \), is a p-series with p=1 which diverges. Hence, the given series \( \sum_{n=1}^{\infty} \sin \frac{1}{n} \) also diverges.