# 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

• Take $f(x)=x$ and $g(x)=2\sin{x}$. We have $f(0)=g(0)=0$, $f'(0)=1<2=g'(0)$ and the derivatives of $f$ and $g$ are continuous, hence there is $\delta>0$ such that $x<2\sin{x}$ for $x\in(0,\delta)$. – user539887 Jul 11 '18 at 12:07
• $$\frac{2}{n\pi}<\sin\left(\frac{1}{n}\right)<\frac{1}{n} \, .$$ – Riemann Jul 11 '18 at 12:08
• @user539887 this is too gud. Thank you – Magneto Jul 11 '18 at 12:27

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 $0<x=\frac1n \le 1$.

• Thank you. Is there any other way to prove riemann sir's identity as given above. From same manner, i see that 2x/pi < sin x if $0<x< \sqrt(6(1-2/ \pi)) = 1.4766$ – Magneto Jul 11 '18 at 12:26
• @Magneto Yes that's also a good way! – gimusi Jul 11 '18 at 12:28

For $0\le x\le\frac\pi2$, in this answer it is shown that $\tan(x)\ge x$. Therefore, \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\frac{\sin(x)}x &=\frac{\cos(x)}{x^2}\,(x-\tan(x))\\ &\le0 \end{align}

Thus, $\frac{\sin(x)}x$ is decreasing on $\left[0,\frac\pi2\right]$, and since $\lim\limits_{x\to0}\frac{\sin(x)}x=1$, $$\frac2\pi\le\frac{\sin(x)}x\le1$$

# Proof

Denote $\dfrac{1}{n}=x,(n=1,2,\cdots)$. Thus, $0<x \leq 1<\dfrac{\pi}{2}$. This shows that the angle with the radian $x$ is a first-quadrant angle on the unit circle. As the figure shows, $\angle AOB=x$，which is bisected by $OD$. Thus,$$|\widehat{AEB}|=x,~~~|\widehat{AE}|=\frac{1}{2}x,~~~ |BC|=\sin x.$$

Notice that $$|\widehat{AEB}|>|AB|>|BC|>|BG|=|DA|>|\widehat{AE}|.$$ Thus, we may have $$x>\sin x>\frac{1}{2}x,$$which is desired.

• By the way, who can tell me how to input the arc sign with Latex here, please? – mengdie1982 Jul 11 '18 at 12:58
• Not sure if you meant for this but; \widehat{ABC} gives $\widehat{ABC}$ – Holo Jul 11 '18 at 13:24
• @Holo Thanks. but i need a smooth "arc" not an angle-like form. The code "\overarc" can't work here. – mengdie1982 Jul 11 '18 at 13:30
• sadly, this side works with Mathjax and not LaTeX, Mathjax support some symbols of LaTeX but not all of them, and the only ways(I know) to create the symbol you want is through the LaTeX package "arcs" with the comment "\overarc" or through the package "fourier" with the comment "\widearc", both packages are not supported in Mathjax – Holo Jul 11 '18 at 13:53

A negative value of derivative $f'$ at some point $x_0$ does not necessarily leads to negative value of the original function $f(x_0)$. It may however indicate that the function is approaching a value from above asymptotically, if the limit is left-limit. If it is a right-limit, everything is reversed.