If $f$ is a continuous periodic fonction on $[0,1]$ and $a\not\in\mathbb{Q}$, the Weyl's equidistribution theorem states that $$\frac{1}{n}\sum_{k=0}^{n1}f(ak)\rightarrow \int_0^1 f(x)dx.$$ Can we say something about the speed of convergence ? Thank you for your help !

As a complement to Gerry Myerson answer, you can bound the discrepancy $D_N$ using the ErdosTuran inequality $$ D_{N} \leq \frac{\log 2}{\pi (H + 1)} + \frac{1}{\pi N} \sum_{h = 1}^{H} \frac{1}{h} \bigg  \sum_{n = 1}^{N} \exp(2\pi i h x_n) \bigg  $$ with $H$ arbitrary. The ErdosTuran inequality is a reasonably easy consequence of Poissonsummation. To get nice constants as above one uses BeurlingSelberg majorants. See: http://www.math.northwestern.edu/~mlerma/papers/charla/appl_of_ef.pdf and the references therein (in particular the survey of Valeer in the Bulletin). The result cited above is Theorem 3.2 in the link. 


Koksma's inequality says $$\leftN^{1}\sum_1^Nf(x_n)\int_0^1f(t)\,dt\right\le V(f)D^*_N$$ where $V(f)$ is the variation of $f$ (we're assuming $f$ is of bounded variation) and $D^*_N$ is the "discrepancy" of $\{x_1,\dots,x_N\}$. See Kuipers and Niederreiter, Uniform Distribution of Sequences, for this and more. 


As you see from the above two answers the rate of convergence will depend on the diophantine nature of $\alpha$. Indeed $$ \sum_{n = 1}^{N} \exp(2\pi i h \alpha n)\ll \min(N,1/h \alpha) $$ where $\xi$ is the distance of $\xi$ to the nearest integer. Therefore the speed of convergence will be bounded by $$ V(f) \cdot \bigg ( \frac{1}{H} + \frac{1}{N} \sum_{h = 1}^{H} \frac{\min(N,1/h \alpha)}{h} \bigg ) $$ and thus depend highly on the algebraic nature of $\alpha$. Case: Algebraic numbers For example if $\alpha$ is algebraic then by Roth's theorem there are only finitely many solutions to $q \alpha < q^{1  \varepsilon}$, therefore for any given $\varepsilon$ we will have $ h \alpha  > c(\varepsilon) h^{1  \varepsilon}$ therefore $1 / h \alpha \leq c'(\varepsilon) h^{1 + \varepsilon}$. Therefore we get the following bound for the discrepancy $$ \ll_{\varepsilon} V(f) \cdot \bigg ( \frac{1}{H} + \frac{H^{1 + \varepsilon}}{N} \bigg ) $$ provided that $H \ll_{\varepsilon} N^{1  \varepsilon}$. Choosing $H = \sqrt{N}$ we conclude that the discrepancy for algebraic numbers is bounded by $$ \ll_{\varepsilon} V(f) N^{1/2 + \varepsilon} $$ Case: Transcendental numbers In general you can pick a transcendental number that will make convergence as slow as one wishes. If you are interested in specific transcendental numbers such as $\pi$ and $e$ then you might use results on the measure of transcendence to obtain a rate of convergence, much in the same manner as done above for algebraic numbers. 

