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 .

Question

Aurora Alvarado · Accepted Answer

This series is convergent, because #\sum_{n=1}^\infty \frac{1}{n^2}# is a convergent #p#-series (#p>1#).

There are a couple of things you need to know for this problem: 1) that #\lim_{x\to 0}\frac{\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 #\sum_{n=1}^\infty a_n# and #\sum_{n=1}^\infty b_n# with postive terms, if #\lim_{n\to \infty}\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 #\sum_{n=1}^\infty \frac{1}{n^p}# converges iff #p>1#. See here for details. Now, we are looking at the case #a_n=\sin^2\(\frac{1}{n})# and #b_n=\frac{1}{n^2}#. Then #\lim_{n\to \infty} \frac{a_n}{b_n}=\lim_{n\to\infty}\frac{\sin^2\(\frac{1}{n})}{\frac{1}{n^2}}=\lim_{n\to \infty}(\frac{\sin(\frac{1}{n})}{\frac{1}{n}})^2#. Let #t=\frac{1}{n}#. Then #\lim_{n\to \infty}t=0# so that #\lim_{n\to\infty}\frac{\sin\(\frac{1}{n})}{\frac{1}{n}}=\lim_{t\to 0}\frac{\sin t}{t}=1# and thus #\lim_{n\to \infty} \frac{a_n}{b_n}=1^2=1>0#. By the Limit Comparison Test, #\sum_{n=1}^\infty a_n# and #\sum_{n=1}^\infty b_n# behave similarly (in terms of convergence). But we know that #\sum_{n=1}^\infty \frac{1}{n^2}# is a convergent #p#-series (#p>1#), so that #\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \frac{\sin^2(\frac{1}{n})}{\frac{1}{n^2}}# is convergent.

Ezra Adams · Answer

undefined To test the convergence of the series $ \sum_{n=1}^\infty \sin^2\left(\frac{1}{n}\right) $ using the limit comparison test with $ c_n = \frac{1}{n^2} $:

1. Calculate the limit as $ n $ approaches infinity of $ \frac{\sin^2\left(\frac{1}{n}\right)}{\frac{1}{n^2}} $.
2. If the limit is a finite positive number, then the series $ \sum_{n=1}^\infty \sin^2\left(\frac{1}{n}\right) $ converges or diverges in the same manner as the series $ \sum_{n=1}^\infty \frac{1}{n^2} $.
3. Since $ \sum_{n=1}^\infty \frac{1}{n^2} $ is a convergent $ p $-series with $ p = 2 $, if the limit from step 1 is finite and positive, then the original series also converges.

The limit comparison test is based on the fact that if $ \lim_{n \to \infty} \frac{a_n}{b_n} = L $, where $ L $ is a finite positive number, then the series $ \sum a_n $ and $ \sum b_n $ either both converge or both diverge.