Thanks to s.harp answer I edited my question
Let $f_n \in C^{\infty}(\mathbb{R}, \mathbb{R}) $ be sequence of function that converges uniformally to a function $ f \in C^{\infty}(\mathbb{R}, \mathbb{R})$ and suppose that we can expand $f_n$ in taylor series
$$ f_n(x)=f_n(a)+\frac{f_n^{\prime}(a)}{1 !}(x-a)+\frac{f_n^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f_n^{(3)}(a)}{3 !}(x-a)^{3}+\cdots $$
Now suppose that the derivatives $f_n^{(i)}$ converges also uniformally to $v^i$.
Is it true that $$ f(x)=f(a)+\frac{v^{1}(a)}{1 !}(x-a)+\frac{v^{2}(a)}{2 !}(x-a)^{2}+\frac{v^{3}(a)}{3 !}(x-a)^{3}+\cdots $$ How can we proof that?
No. Take: $$f_n(x)= \exp\left(\frac{-1}{x^2+1/n}\right)$$ which is analytic in some neighbourhood of $0$. In some neighbourhood of $0$ this also converges uniformly to $\exp(-1/x^2)$. The derivatives also all uniformly converge to the derivatives of $\exp(-1/x^2)$ in such a neighbourhood of $0$.
But $\exp(-1/x^2)$ is not analytic in any neighourhood of $0$, meaning that the Taylor series does not converge to $f$.