Does the series $\sum\sin^2(\frac{1}{n})$ diverge or converge?
I tried the comparison test. We have $\sin(n) \leq n$ for all natural numbers $n$, so $\sin(\frac{1}{n}) \leq \frac{1}{n}$, and $\sin^2(\frac{1}{n}) \leq \frac{1}{n^2}$ for all natural numbers $n$.
Note that the series $\sum\frac{1}{n^2}$ converges, so the series $\sum\sin^2(\frac{1}{n})$ converges by the comparison test.
What have I messed up on in my working and explanations?