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.
