How to show that $\sin^2(1/n)\leq 1/n^2, \forall n\in\mathbb{N}?$ I'm trying to prove that $\sum_{n=1}^\infty \sin^2(1/n)$ is convergent, by using the comparison test. 
I hypothesize that the sequence defined by $b_n:=1/n^2$ is always larger than $\sin^2(1/n)$. But when I do induction on $n$, the inequality isn't clear.
 A: Your hunch is correct, for all $x\in\mathbb{R}_{\geqslant 0}$, $\sin(x)\leqslant x$ as it can be shown using the derivative of $x\mapsto \sin(x)-x$ or the Taylor formula with integral remainder.
A: When deriving the derivative of $\sin$, you likely found the bounds
$$\cos(x)<\frac{\sin(x)}x<1\quad\forall~0<|x|<\pi/2$$
From this, you can easily see that
$$\sin(x)<x\quad\forall~0<x\le1\\\implies\sin^2(1/n)<1/n^2\quad\forall~1\le n$$

If you don't happen to remember this off the top of your head, it may be more intuitive to see that if $f''(x)<0~\forall~x\in(a,b)$, then $f(x)$ is bounded on that interval by it's tangent line and secant lines:
$$f(a)+\frac{f(a)-f(b)}{a-b}(x-a)<f(x)<f(a)+f'(a)(x-a)\quad\forall~a<x<b$$
Particularly,
$$\sin(x)<\sin(0)+\cos(0)x=x\quad\forall~0<x<\pi$$
A: Hint Use the Limit Comparison Test instead:
$$\lim_n \frac{\sin^2(1/n)}{1/n^2} =\left( \lim_n \frac{\sin(1/n)}{1/n} \right)^2=\left( \lim_{x_n \to 0} \frac{\sin(x_n)}{x_n} \right)^2=1$$
A: If you are allowed to use the limit comparison test, I recommend it.  Use your same $b_n$.  Hint to get you started:
$$\lim_{n\to+\infty} \frac{a_n}{b_n} = \lim_{n\to+\infty} \frac{\sin^2(1/n^2)}{1/n^2}$$
Now make the substitution $m = 1/n$.  What does $m$ approach as $n \to +\infty$?  The result of this substitution should be a familiar limit from Calc 1.
A: From the geometric definition of $\sin x:$
For $0<x<\pi /2,$ take triangle $AOB$ with $OA=OB=1$ and $\angle AOB=x.$ Let D lie on $OA$ with $DB\perp OA.$ The arc of the circle centered at $O,$ of radius $1,$ from  $A$ to  $B,$ has length $x,$ which is greater than the straight-line distance $AB.$ So $$\sin x=DB<\sqrt {DB^2+DA^2}\;=AB<x.$$
Since $\sin (-x)=-\sin x,$ therefore $|\sin x|<|x|$ for $0<|x|<\pi /2.$ 
For $|x|\geq \pi /2$ we have $|x|>1\geq |\sin x|.$
And of course for $x=0$ we have $|\sin x|=0=|x|.$
A: We have: $$|\sin x|=|\int_0^x\cos t dt|\le\int_0^x|\cos t| dt\le x$$
take $x=1/n$
Hence, $$\sum_{n=1}^\infty \sin^2(\frac1n)\le \sum_{n=1}^\infty\frac{1}{n^2}<\infty$$
