Show that $|\sin\frac{1}{n^2}|<\frac{1}{n^2}$, $n=0, 1, 2, \dots$ As part of showing that
$$
\sum_{n=1}^\infty \left|\sin\left(\frac{1}{n^2}\right)\right|
$$
converges, I ended up with trying to show that
$$
\left|\sin\left(\frac{1}{n^2}\right)\right|<\frac{1}{n^2}, \quad n=1, 2, 3,\dots
$$
since I know that the sum of the right hand side converges. But I can't show this. I've tried searching but I haven't been able to find anything.
What I've tried is that firstly, the absolute values are not needed since $\sin x>0$ if $0<x<1$. I rearranged a little bit:
$$
\sin\left(\frac{1}{n^2}\right)-\frac{1}{n^2}<0
$$
and the derivative is
$$
\frac{2}{n^3}\left(1-\cos\left(\frac{1}{n^2}\right)\right)> 0
$$
so my idea of showing that it is decreasing and negative for the first $n$ wouldn't work. 
How can I show this? Help is appreciated. 
Edit: Maybe I should add that I'm not completely sure it is true, but I tried it numerically and it seems like it. 
 A: Consider $f(x)=\sin x -x\Rightarrow f'(x)=\cos x-1 < 0 \text { for x > 0 } \Rightarrow f(x)\text{ is strictly decreasing function}$
But then, $f(0)=0$ which would imply $f(x)< 0 $ for $x > 0$ i.e., $\sin x < x$ for $x >0$ 
Thus, $\sin \big(\frac{1}{n^2}\big)< \frac{1}{n^2}$
Does this imply $|\sin\big(\frac{1}{n^2}\big)|< \frac{1}{n^2}$
A: It's actually very straightforward to show that $\sin x\leq x$ for $0<x<1$ which is all you have to do...
You can do it by definition using a unit circle.
You can do it by noting that $\sin 0=0$ and $\cos x<1$ here (i.e. $\sin x$ grows slower than $x$ here).
A: If $0 < x < \pi/2$, $\sin(x)$ is the $y$--coordinate of the point on the unit circle distance $x$ via the circle.  That is the shortest distance of any path from the point $P$, $(\cos(x), \sin(x))$ to the $x$--axis.   Since $x$ describes the length of a path from $P$ to the $x$--axis that is not a straight line, we have $\sin(x) < x$.  Draw of picture of this to see it nicely.
A: Hint: $|\sin(t)| \le |t|$ for real $t$, with equality only at $0$.  
