convergence of $ \sum_{k=1}^\infty \sin^2(\frac 1 k)$ 
convergence of $ \sum_{k=1}^\infty \sin^2(\frac 1 k)$

How can I do this? Should I use the Ratio Test (I tried this but it started getting complicated so I stopped)? Or the Comparison test(what should I compare it to?)?

With equivalents:
$\sin^2\dfrac1k\sim_\infty \dfrac1{k^2}$.
The series $\displaystyle\sum\dfrac1{k^2}$ converges absolutely, hence $\displaystyle\sum\sin^2\dfrac1k$ converges absolutely.
I recall $f(x)\sim_\infty g(x)$, for two functions defined in a neighbourhood of $\infty$ roughly means
$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=1$$
To take into account that $g(x)$ can be equal to $0$, there is a technical definition:
$$f(x)-g(x)=_{\infty}o\bigl(g(x)\bigr),$$
which is equivalent to the  previous definition in case $g(x)\neq 0$ in some neighbourhood of $\infty$.
