Function with roots at $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ...$? I am searching for a function $~f : \mathbb{R}\rightarrow \mathbb{R}$ with
$$ f\left(\frac{1}{k}\right) = 0 ~\forall k\in\mathbb{N}. $$
This function has to be smooth! 
First approach: I have tried $f(x) = \sin(\pi/x)$. This works, however, this function is not smooth (as far as I am concerned?).
Context: The overall aim is to find two functions $f,g$ such that $f\neq g$ but $f(1/k) = g(1/k) ~\forall k\in\mathbb{N}$. 
EDIT: Smooth means $f\in C^\infty$. $~f$ has to be at least defined on $(0,1]$. 

Using the function $e^{-1/x^2}$ for $x \ne 0$, and $0$ if $x = 0$, we get a function which is $C^{\infty}$ and supported on $[0, \infty)$. Multiplying this by a shift and flip of the function, we can easily construct a function which is $C^{\infty}$ and supported only on $[0, 1]$, giving a bump function.
Summing translates and dilates of thus localized to the intervals $[1/2, 1]$, $[1/3, 1/2]$, $[1/4, 1/3]$ and so on gives an example of a function whose zeros in $[0, 1]$ are only at the points $1, 1/2, 1/3, ...$.
