Proof on $F(x)=\sum_{n=1}^{\infty}f\left(\frac{x}{n}\right)$ uniform convergence and differentiability Let $f$ be a function of $C^{\infty}$ class, such that $f(0)=0=f'(0)$. Prove that if $x\in\mathbb{R}$ and $$F(x)=\sum_{n=1}^{\infty}f\left(\frac{x}{n}\right)\ ,$$
then $F(x)\in\mathbb{R}$ and $F$ is differentiable infinitely many times.
I'm aware of merely one theorem that may be helpful in this case, which is that if $\left(f_n\right)$ is a function sequence that is convergent for some $x_0$ and $\left(f'_n\right)$ is uniformly convergent, then $\left(f_n\right)$ is also uniformly convergent (...). 
We know that $F(0)$ is finite. Then for $$G(x)=\sum_{n=1}^{\infty}\frac{1}{n}f'\left(\frac{x}{n}\right)$$ we also have $G(0)<\infty$. If I could show $G$ is uniformly convergent, it would mean $F$ is uniformly convergent and $F'=G$.
Now, I'm not sure if what I'm about to do is correct, but I'd like to use the dfinition of near uniform convergence. $f\in C^{\infty}$, so $f'\in C^{\infty}$. Every fucntion continuous on closed interval is uniformly continuous, and in result, is Lipschtiz. So for $x\in[a,b]$ and some $M_1,M_2\in\mathbb{R}$
$$\sum_{n=1}^{\infty}\frac{1}{n^2}f''\left(\frac{x}{n}\right)-\frac{1}{n^2}f'\left(\frac{x}{n}\right)\leq \sum_{n=1}^{\infty}\frac{M_2-M_1}{n^2} < \infty$$
This means, that by Weierstrass theorem $G'$ is uniformly convergent $\Rightarrow G$ is u.c., which in turn basing on above-mentioned theorem again, gives us uniform convergence of $F$.
Mmm, as I look at this proof, it seems dubious... Moreover, I didn't include differentiability of $F$. Please, help guys.
 A: Because $\lim_{x\to0}\frac{f(x)}{x^2}$ exists (use l'Hôpital), it follows that $|f(x)|<Cx^2$ for small $x$. Because in dealing with convergence one can always delete finitely many terms, we start with some big $N$, so $\frac{x}{n}$ is small, to use this inequality. It follows that $F(x)<\infty$:
$$\left|\sum_{n=N}^\infty f\left(\frac{x}{n}\right)\right|\leq Cx^2\sum_{n=N}^\infty \frac{1}{n^2}<\infty$$
Because $\lim_{x\to0}\frac{f'(x)}{x}$ exists, it follows that $|f'(x)|<C|x|$ for small $x$. Now we see that $F'(x)<\infty$.
$$\left|\sum_{n=N}^\infty \frac{1}{n}f'\left(\frac{x}{n}\right)\right|\leq C\sum_{n=N}^\infty \frac{|x|}{n^2}<\infty$$
Now we want to show that $F^{(k)}(x)<\infty$, $k\geq2$. Here we use that $|f^{(k)}(x)|<C$ for small $x$
$$\left|\sum_{n=N}^\infty\frac{1}{n^k}f^{(k)}\left(\frac{x}{n}\right)\right|\leq C\sum_{n=N}^\infty\frac{1}{n^k}$$
In your post you mention uniform convergence, but that is not required to answer the question. If we for example take $f(x)=x^2$, then $F(x)=\frac{\pi^2}{6}x^2$, but the difference between every partial sum and the limit function is not bounded.
