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Let $\{f_i\}_{i \in \mathbb{N}}$ be a sequence of differentiable real-valued functions in $\mathbb{R}$.
Assume that for any $x \in \mathbb{R}$, the sum $N(x) = \sum\limits_{i=0}^\infty f_i(x)$ is convergent. Is there a general theorem that states that $N(x)$ is also a differentiable function?
Yes: if the series $\sum_{i=0}^\infty f_i'(x)$ converges uniformly to $g$, then $N$ is differentiable and $N'=g$.
In general the sum is not differentiable even if $\sum_{n=1}^\infty f_n$ converges uniformly.
Example from @zhw. here:
Consider the smooth functions $f_n : \mathbb{R} \to \mathbb{R}$ defined as $f_n(x) = \sqrt {x^2+\frac1n}$.
For any $x\in \mathbb{R}$ we have
$$0\le f_n(x) - |x| =\sqrt { x^2+\frac1n }-\sqrt {x^2} =\frac{1/n}{ \sqrt {x^2+\frac1n}+\sqrt {x^2} } \le \frac{\frac1n}{\frac1{\sqrt n}} = \frac{1}{\sqrt n} $$
so $f_n \xrightarrow{n\to\infty} |\cdot|$ uniformly on $\mathbb{R}$. Every sequence can be turned into a series so we have:
$$f_1 + \sum_{n=1}^{\infty}(f_{n+1}-f_n) \xrightarrow{n\to\infty} |\cdot|$$
however the absolute value $|\cdot|$ is clearly not differentiable at $0$.
