Consider the series
![\[ \sum_{n=1}^{\infty} \left( \sin \frac{1}{n} \right)^{\frac{3}{2}}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5f980bddba22d45866d55b7544f285f1_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 since it is of the form 
for 
. Using the limit comparison test we have
![\begin{align*} \lim_{n \to \infty} \frac{a_n}{b_n} &= \lim_{n \to \infty} \frac{\left( \sin \frac{1}{n} \right)^{\frac{3}{2}}}{\left( \frac{1}{n} \right)^{\frac{3}{2}}} \\[9pt] &= \lim_{n \to \infty} \left( \frac{\sin \frac{1}{n}}{\frac{1}{n}} \right)^{\frac{3}{2}} \\[9pt] &= 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-c04a76ee65d0054036e0f427aced79e9_l3.png "Rendered by QuickLaTeX.com")
Hence, the series and 
both converge or both diverge. Since converges we have establish the convergence of
![\[ \sum_{n=1}^{\infty} \left( \sin \frac{1}{n} \right)^{\frac{3}{2}}. \qquad \blacksquare\]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ab63536243b271b9173080106df3cc9c_l3.png "Rendered by QuickLaTeX.com")
You need to prove $a_n>0$ for all $n>0$ (which, of course, if trivial) to use the Comparison Test.