<s>[INST] How is $\frac{1}{2n} \leq \sin (\frac{1}{n}) \leq \frac{1}{n} $ How is $\frac{1}{2n} \leq \sin (\frac{1}{n}) \leq \frac{1}{n} $
I know that $\sin \theta \leq \theta, \theta$ very small
But if take $ f(x) = \sin (\frac{1}{x}) -  \frac{1}{2x}, f'(x) = \frac{-1}{x^2} \cos (\frac{1}{x}) +  \frac{1}{2x^2} $
But if i take $x=\frac 4\pi, x \in (0, \pi/2), $ i am getting $f'(x) < 0$ which should be other way around. Am I missing something? I got this doubt while reading Does $\sum_{n=1}^\infty(-1)^n \sin \left( \frac{1}{n} \right) $ absolutely converge?
If i say since $\sin (x)$ converges to $x$, i will have $\sin x$ values slightly less than $x$ and slightly more than $x$. But it depends on whether $f(x)$ is increasing/decreasing (local maxima or local minima)
Pls clarify
 [/INST] We know that for $x=\frac1n>0$ 
$$x-\frac16x^3\le \sin x \le x$$
[ refer to Proving that $x - \frac{x^3}{3!} < \sin x < x$ for all $x>0$ ]
and
$$\frac12x\le x-\frac16x^3 \iff \frac12x-\frac16 x^3\ge 0 \iff x^2\le 3 \iff0<x<\sqrt 3$$
which is true since $0<x=\frac1n \le 1$.
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