The problem is to calculate a limit $$ \lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n \Big\{\frac{k}{\sqrt{3}}\Big\} $$ where {$\cdot$} is a fractional part. I believe that this limit is equal to $\frac{1}{2}$, but I do not have a rigorous proof. The only idea that comes to my mind is the following one. It can be shown that the set $$ A=\bigg\{ \Big\{\frac{k}{\sqrt{3}}\Big\} : k\in \mathbb{N}\bigg\} \subset [0,1] $$ is dense and equidistributed. That's why the mean value represented by the limit is equal to the mean value of the uniform distribution $U(0,1)$ which is equal to $1/2$. How can I make the proof rigorous?
Any other approaches are appreciated. Probably one can apply law of big numbers to calculate this limit.
You may use a rescaled version of Khinchin's equidistribution theorem:
$$ \lim_{n\to +\infty}\frac{1}{n \sqrt{3}}\sum_{k=1}^{\left\lfloor n\sqrt{3}\right\rfloor} f\left((x+ka)\!\!\!\!\pmod{\sqrt{3}}\right) = \frac{1}{\sqrt{3}}\int_{0}^{\sqrt{3}}f(y)\,dy. $$ We have $\left\{\frac{k}{\sqrt{3}}\right\}=\frac{1}{\sqrt{3}}\left(k\!\!\pmod{\sqrt{3}}\right)$, hence the limit is given by: $$ \frac{1}{3}\int_{0}^{\sqrt{3}}y\,dy=\color{red}{\frac{1}{2}}$$ as expected (pun intended, now).
Anyway, since we are dealing with a trivial case of Khinchin's theorem, we may just expand $\{x\}-\frac{1}{2}$ as a Fourier sine series and use termwise integration to prove just the same.