Find $\sum f_n(x)$ Problem
For non-negative integer $n$, define
$$f_n\colon [0, 1]\to\mathbb{R}, \quad f_n(x)=\int_0^x f_{n-1}(t)dt\quad (n>0),\quad f_0(x) = e^x$$
Find $g(x)=\displaystyle\sum_{n=0}^\infty f_n(x)$

My attempt
Compute $f_n(x)$ for some $n$ :
$f_0=e^x$
$f_1=e^x-1$
$f_2=e^x-x-1$
$f_3=e^x-\frac{x^2}{2}-x-1$
so, $f_n$ has series expansion until $(n-1)$-th degree of $e^x$.
My conjecture is :
$$g(x)= \sum_{n=0}^\infty f_n(x)=e^x + \sum_{n=1}^\infty f_n(x) = e^x$$
But I don't know how to justify this one.
 A: If you find a way to justify that we can differentiate $g$ termwise, we get:
$$g'(x) = \sum_{n\ge 0} f_n'(x) = f'_0(x) + \sum_{n\ge 1} f_n'(x) = f'_0(x) + \sum_{n\ge 1} f_{n-1}(x)  \implies$$
$$g'(x) = f'_0(x) + \sum_{n\ge 0} f_{n}(x) = e^x + g(x)$$
We get $g(x) = (x+1) e^x$
A: A theorem states

If $(f_{n})$ is a sequence of differentiable functions on $[a,b]$ such that $\lim\limits_{n\to \infty }f_{n}(x_{0})$ exists (and is finite) for some $x_{0}\in [a,b]$ and the sequence $(f^\prime_{n})$ converges uniformly on $[a,b]$, then $(f_{n})$ converges uniformly to a function $f$ on $[a,b]$, and $f^\prime(x)=\lim\limits_{n\to \infty }f^\prime_{n}(x)$ for $x\in [a,b]$.

Now using Taylor's Theorem, for $x \in [0,1]$ and $n \ge 0$, it exists $\xi \in (0,x)$ such that
$$f_{n+1}(x) = e^x - P_n(x) = \frac{e^\xi}{(n+1)!} x^{n+1}$$ where
$$P_n(x) = \sum_{k=0}^n \frac{x^k}{k!}.$$ Therefore $\vert f_{n+1}(x) \vert \le \frac{e}{(n+1)!}$. Which implies that $\sum f^\prime_n$ converges uniformly according to Weierstrass M-test. As $f_n(0) = 0$ for $n \ge 1$, we can apply the theorem mentioned above.
And conclude that
$$g(x)=(x+1) e^x$$ as stated in JustANoob answer.
