how could i evaluate the derivative of
$ f(x)= \sum_{n=1}^{\infty} [\frac{x}{n}] $
here $ [x] $ is the integer part so $ u(x)=x-[x] $ is the fractional part of the number
form distibution theory i believe that the derivative is somehow
$ f'(x)= \sum_{n=1}^{\infty}\sum_{m=-\infty}^{\infty}\delta (x-nm) $
but i'm not completely sure
The series converges in the sense of distributions - setting
$$f_k(x) = \sum_{n = 1}^k \biggl\lfloor \frac{x}{n}\biggr\rfloor,$$
the sequence $(f_k)$ is monotonically increasing on $[0,+\infty)$ and decreasing on $(-\infty,0)$, so
$$\lim_{k\to\infty} \int_{\mathbb{R}} f_k(x)\varphi(x)\,dx$$
exists for every test function $\varphi$. Therefore we can interchange differentiation (in the sense of distributions) and summation. Thus one needs to find the distributional derivative of
$$g_n \colon x \mapsto \biggl\lfloor \frac{x}{n}\biggr\rfloor$$
for $n \geqslant 1$. Since $g_n$ is constant on each interval $[mm, (m+1)n)$ and has jumps of height $1$ at each point $mn$, indeed - in the sense of distributions -
$$g_n' = \sum_{m\in \mathbb{Z}} \delta_{mn},$$
and your expression for the distributional derivative of $f$ is correct.