Examine whether $\sum_{n=1}^\infty{\sin^2\left(\frac{1}{n}\right)}$ converges or not I have the series $\sum_{n=1}^\infty{\sin^2\left(\frac{1}{n}\right)}$ and I'm trying to examine whether it converges or not.
My Attempts:


*

*I first tried finding whether it diverges by checking if $\lim_{n\to\infty}{\sin^2\left(\frac{1}{n}\right)} \ne 0$.


$$
\lim_{n\to\infty}{\sin^2\left(\frac{1}{n}\right)}=
\lim_{n\to\infty}{\sin\left(\frac{1}{n}\right)}\cdot\lim_{n\to\infty}{\sin\left(\frac{1}{n}\right)}=0
$$


*Since I didn't get a confirmation from the first try, I then tried the d'Alembert's Criterion which didn't get me very far.


$$
\frac{a_{n+1}}{a_n}=
\frac{\sin^2\left(\frac{1}{n+1}\right)}{\sin^2\left(\frac{1}{n}\right)}=
\frac{
   -\dfrac{2\cos\left(\frac{1}{n}\right)\sin\left(\frac{1}{n}\right)}{n^2}
}{
   -\dfrac{2\cos\left(\frac{1}{n+1}\right)\sin\left(\frac{1}{n+1}\right)}{\left(n+1\right)^2}
}=
\frac{
   \cos\left(\frac{1}{n}\right)\sin\left(\frac{1}{n}\right)\left(n+1\right)^2
}{
   \cos\left(\frac{1}{n+1}\right)\sin\left(\frac{1}{n+1}\right)n^2
}=\ ...
$$


*Finally, I tried Cauchy's Criterion, but I didn't get any conclusive result either.


$$
\sqrt[n]{a_n}=
\sqrt[n]{\sin^2\left(\frac{1}{n}\right)}=
\sin^{\frac{2}{n}}\left(\frac{1}{n}\right)=\ ...
$$
Question:
I've been thinking for while of using the Comparison Test, but I'm not sure which series to compare mine to. How can I examine whether the series converges or not?

Observe that
$$
\sin^2(1/n)\sim\frac{1}{n^2}.
$$
Indeed,
$$
\lim_{n\to\infty}\frac{\sin^{2}(1/n)}{1/n^2}=\left[\lim_{n\to\infty}\frac{\sin(1/n)}{1/n}\right]^2=1
$$
where in the last equality we used the well known limit
$$
\lim_{x\to 0}\frac{\sin x}{x}=1.
$$ 
The limit comparison test implies that your series converges.
