convergence of $f_n(x)=f(x+\frac{1}{n})$ if $f:\mathbb{R}\to \mathbb{R}$ is only a function. Let $f:\mathbb{R}\to \mathbb{R}$ a function only. I and let $f_n(x)=f(x+\frac{1}{n})$. I need show if $f_n$ converges uniform and puntually but I do not understand the question. 
I can examples of functions so that $f$ converges or not converges in general?
because $f$ not have conditions.

Not sure I fully understand you here, but you're right that if there are no conditions on $f,$ then there's not much you can say. For instance, if we fix an $x,$ then we can choose an $f$ such that $f(x+\frac{1}{n})$ is any sequence we wish ($f$ is arbitrary, so just define $f(x+1/n) = a_n$ for all points $x+1/n$ and define it to be whatever you want everywhere else). It's harder to show that there is an $f$ such that $f(x+1/n)$ does not converge for all $x$ (but I would expect this to be case for a generic $f$).
If $f$ is continuous, we're in a vastly different situation since $f(x+1/n)$ converges to $f(x)$ for all $x.$ However even here the convergence needn't be uniform. For instance if $f(x) = e^x$ then $$\sup_{x\in \mathbb R}(f(x+1/n)-f(x)) = \sup_{x\in \mathbb R}(e^x(e^{1/n}-1)) = \infty$$
