# How do you test this series? sum for n=1 to infinity sin^2(1/n) convergence , by using limit comparison test with cn=1/n^2 .

Jun 7, 2018

This series is convergent, because $\setminus {\sum}_{n = 1}^{\setminus} \infty \setminus \frac{1}{{n}^{2}}$ is a convergent $p$-series ($p > 1$).

#### Explanation:

There are a couple of things you need to know for this problem: 1) that $\setminus {\lim}_{x \setminus \to 0} \setminus \frac{\setminus \sin x}{x} = 1$. You can refer to this for a geometric argument from scratch, or use the Rule of de l'Hospital if you've seen it.

2) the limit comparison test, stating that for two series $\setminus {\sum}_{n = 1}^{\setminus} \infty {a}_{n}$ and $\setminus {\sum}_{n = 1}^{\setminus} \infty {b}_{n}$ with postive terms, if $\setminus {\lim}_{n \setminus \to \setminus \infty} \setminus \frac{{a}_{n}}{{b}_{n}} = c$ for $c$ a finite positive number, then either both series converge, or both diverge. See for instance here for details.

3) That $p$-series of the form $\setminus {\sum}_{n = 1}^{\setminus} \infty \setminus \frac{1}{{n}^{p}}$ converges iff $p > 1$. See here for details.

Now, we are looking at the case ${a}_{n} = \setminus {\sin}^{2} \setminus \left(\setminus \frac{1}{n}\right)$ and ${b}_{n} = \setminus \frac{1}{{n}^{2}}$.
Then $\setminus {\lim}_{n \setminus \to \setminus \infty} \setminus \frac{{a}_{n}}{{b}_{n}} = \setminus {\lim}_{n \setminus \to \setminus \infty} \setminus \frac{\setminus {\sin}^{2} \setminus \left(\setminus \frac{1}{n}\right)}{\setminus \frac{1}{{n}^{2}}} = \setminus {\lim}_{n \setminus \to \setminus \infty} {\left(\setminus \frac{\setminus \sin \left(\setminus \frac{1}{n}\right)}{\setminus \frac{1}{n}}\right)}^{2}$.

Let $t = \setminus \frac{1}{n}$. Then $\setminus {\lim}_{n \setminus \to \setminus \infty} t = 0$ so that $\setminus {\lim}_{n \setminus \to \setminus \infty} \setminus \frac{\setminus \sin \setminus \left(\setminus \frac{1}{n}\right)}{\setminus \frac{1}{n}} = \setminus {\lim}_{t \setminus \to 0} \setminus \frac{\setminus \sin t}{t} = 1$ and thus $\setminus {\lim}_{n \setminus \to \setminus \infty} \setminus \frac{{a}_{n}}{{b}_{n}} = {1}^{2} = 1 > 0$.

By the Limit Comparison Test, $\setminus {\sum}_{n = 1}^{\setminus} \infty {a}_{n}$ and $\setminus {\sum}_{n = 1}^{\setminus} \infty {b}_{n}$ behave similarly (in terms of convergence). But we know that $\setminus {\sum}_{n = 1}^{\setminus} \infty \setminus \frac{1}{{n}^{2}}$ is a convergent $p$-series ($p > 1$), so that $\setminus {\sum}_{n = 1}^{\setminus} \infty {a}_{n} = \setminus {\sum}_{n = 1}^{\setminus} \infty \setminus \frac{\setminus {\sin}^{2} \left(\setminus \frac{1}{n}\right)}{\setminus \frac{1}{{n}^{2}}}$ is convergent.