Consider the series
![\[ \sum_{n=1}^{\infty} \frac{\sin \frac{1}{n}}{n}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-9bd046572719a1c0cce20b2d506ab9d1_l3.png "Rendered by QuickLaTeX.com")
Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.
The series converges.
Proof. Let 
. Then, the series 
converges. Using the limit comparison test we have
![\begin{align*} \lim_{n \to \infty} \frac{a_n}{b_n} &= \lim_{n \to \infty} \left( \frac{ \frac{\sin \frac{1}{n}}{n}}{\frac{1}{n^2}} \right) \\[9pt] &= \lim_{n \to \infty} n \sin \frac{1}{n} \\[9pt] &= \lim_{n \to \infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}} \\[9pt] &= 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-de8d4a65516e6c6b42ea13d159439c20_l3.png "Rendered by QuickLaTeX.com")
Hence, 
and both converge or both diverge. Since converges we have established the convergence of
![\[ \sum_{n=1}^{\infty} \frac{\sin \frac{1}{n}}{n}. \qquad \blacksquare\]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-b79b117da6b971b8dd967020542d9785_l3.png "Rendered by QuickLaTeX.com")
The integral goes to infinite, not 1. lim n-> infinite of sin(1/n) / (1/n) goes to infinite because 1/n goes to 0 and since it is in the denominator the whole expression goes to infinite
You should review the basic limit facts