Is there a statement like $\sin\left(\frac1{n^k}\right) < \frac1{n^k} \forall n,k\in \mathbb{N}$ I'm doing some exam review and I thought I might need to use something like this.  Is it true? If not, is there a similar statement, and if so, how can we prove it? 
$$\sin\left(\frac1{n^k}\right) < \frac1{n^k} \forall n,k\in \mathbb{N}$$  
I can't find it on the site or in my textbook, but it seems to be an assumption used quite a bit (or some variation of it).  I'm assuming if it's true it can be improved to say for all real numbers greater than 1? Or possibly even better?  
I tried checking on wolfram, but I'm not sure how to ask a question and make it accept conditions : http://www.wolframalpha.com/input/?i=is+sin%281%2Fn%29+%3C+1%2Fn
 A: Consider $f(x)=\sin(x)-x$. Note that $f(0)=0$, and that $f'(x)=\cos(x)-1\le 0$, so $f$ is decreasing, i.e., $f(x)\le0$ for all $x\ge0$, or $\sin(x)\le x$. In particular, $$ \sin\left(\frac1{n^k}\right)\le\frac1{n^k}.$$
In fact, the inequality is strict, since otherwise $t=1/n^k$ satisfies $f(t)=0$. But then $f(0)=f(t)$ and, by the mean value theorem, there is an $s$ between $0$ and $t$ with $f'(s)=0$. But $\cos(x)<1$ for all $x\in(0,2\pi)$, and $1/n^k<2\pi$.
A: It is true.  First notice that if $0 < x < \pi/2$, we have $0 < \sin(x) < x$.  This is true by the following argument.  The point $(\cos(x), \sin(x))$ is the point you arrive at by traveling distance $x$ along the unit circle counterclockwise from $(1,0)$.  Hence $\sin(x)$ is the vertical distance from
$(\cos(x), \sin(x))$ to the $x$--axis.  Since this vertical distance is the shortest path from $(\cos(x), \sin(x))$ to the $x$-axis, we have $0 < \sin(x) < x$ for $0 < x < \pi/2$.  
Now take $x = 1/n^k$, and you are done.
A: $\sin(x) < x$ for all $x > 0$, which includes what you just said.  The proof is using the derivative:
Let $f(x) = x - \sin x$.  Then $f'(x) = 1 - \cos x \geq 0$ for all $x$.  Since it's only equal to 0 at one point here and there, $2n\pi$ for $n \in \mathbb{Z}$, we know $f$ is increasing for all $x$.  Since $f(0) = 0$, we see that $f(x) > 0$ for all $x > 0$, which is the same as $x > \sin x$.
A: Look at the function $\frac{\sin(x)}{x} $ on the interval $(0,1)$ and optimize by taking derivatives and such.  You will find that $\frac{\sin(x)}{x}<1$ with the ''maximum'' attained when $x\to 0.$  Now with a little algebra you should get $\sin(x)<x.$
Now observe that $0<\frac{1}{n^k}<1$ for every $n,k\in \mathbb{N}.$  Thus, we can choose $x=\frac{1}{n^k}$ and you will get your desired inequality.  
