## Calculus 8th Edition

To use the Direct Comparison Test, we need a known series $\Sigma_{n=1}^{\infty}b_{n}$ for the purpose of comparison. We will choose $b_{n}=\frac{1}{n}$ The Comparison Test states that the p-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$. Now,$a_{n}=sin(\frac{1}{n})$ and $b_{n}=\frac{1}{n}$ $\lim\limits_{n \to \infty}\frac{a_{n}}{b_{n}}=\lim\limits_{n \to \infty}\frac{sin(\frac{1}{n})}{\frac{1}{n}}$ Let $t=\frac{1}{n}$, also note that $t \to 0$ as $n \to \infty$ $\lim\limits_{n \to \infty}\frac{a_{n}}{b_{n}}=\lim\limits_{t \to 0}\frac{sint}{t}$ Since, the limit is of the form $\frac{0}{0}$ , we can apply L Hospital's Rule $=\lim\limits_{t \to 0}\frac{cost}{1}$ $=cos (0)$ $=1\ne 0 \ne \infty$ The given series diverges by limit comparison test because $\Sigma _{n=1}^{\infty} \frac{1}{n}$ is diverging.