found the limit function of function sequence that define recursively with integrals Find the limit function of the sequence:
at the domain [0,1].
$$f_{1}= \cos x $$
$$ f_{n+1} (x) =   \int_{0}^{x} f_{n}(t) dt$$
I found the first element of the sequence and I had noticed that Taylor series shows up and it converges to $0$. but I don't know how to prove it. 
 A: Note that $\sin x \leq \cos x$ for $0 \leq x \leq \pi/4$. In this interval we have $f_2 \leq f_1$ and we get $f_{n+1} \leq f_n$ by  a simple induction argument. Of course $|f_n(x)| \leq 1$ for all $n,x$ by induction. Hence $f(x)=\lim f_n(x)$ exists for al $x$. By DCT we get $f(x)=\int_0^{x} f(t) dt$. This gives $f'(x)=f(x)$ and hence $f(x)=ce^{x}$. But $f_n(0)=0$ for all $n>1$ so $f(0)=0$ and $c=0$; thus the limit function is $0$ on $[0,\pi /4]$. On $[\pi /4 ,1]$ we have $\sin x \geq \cos x$ and a similar argument works to show that $f(x)=0$ in this case also. 
A: Let's prove by induction that :
\begin{aligned} \left(\forall n\geq 2\right)\left(\forall x\in\left[0,1\right]\right),\ f_{n}\left(x\right)=\frac{1}{\left(n-2\right)!}\int_{0}^{x}{\left(x-y\right)^{n-2}\cos{y}\,\mathrm{d}y} \end{aligned}
The base case is trivial.
Let $ n\geq 2 $, assuming that the statement is true for $ n $, let's prove it holds for $ n+1 : $
\begin{aligned}\left(\forall x\in\left[0,1\right]\right),\ f_{n+1}\left(x\right)&=\int_{0}^{x}{f_{n}\left(t\right)\mathrm{d}t}\\ &=\frac{1}{\left(n-2\right)!}\int_{0}^{x}{\int_{0}^{t}{\left(t-y\right)^{n-2}\cos{y}\,\mathrm{d}y}\,\mathrm{d}t}\\ &=\frac{1}{\left(n-2\right)!}\int_{0}^{x}{\int_{y}^{x}{\left(t-y\right)^{n-2}\cos{y}\,\mathrm{d}t}\,\mathrm{d}y}\\ \left(\forall x\in\left[0,1\right]\right),\ f_{n+1}\left(x\right)&=\frac{1}{\left(n-1\right)!}\int_{0}^{x}{\left(x-y\right)^{n-1}\cos{y}\,\mathrm{d}y}\end{aligned}
Thus, the statement holds for every $ n\geq 2 \cdot $
Now, thanks to the factorial, we'll get $ 0 $ as a limit for $ f_{n} $. Proof :
\begin{aligned} \left(\forall n\geq 2\right)\left(\forall x\in\left[0,1\right]\right),\ \left|f_{n}\left(x\right)\right|&\leq\frac{1}{\left(n-2\right)!}\int_{0}^{x}{\left|\left(x-y\right)^{n-2}\cos{y}\right|\mathrm{d}y}\\&\leq\frac{1}{\left(n-2\right)!}\int_{0}^{x}{\left(x-y\right)^{n-2}\,\mathrm{d}y}=\frac{x^{n-1}}{\left(n-1\right)!}\\ &\leq\frac{1}{\left(n-1\right)!} \end{aligned}
Thus, $ \left(f_{n}\right)_{n} $ converges $\left(\underline{\textbf{normally}}\right)$ to $ 0\cdot$
$\left(\textrm{Because not only }\frac{1}{\left(n-1\right)!}\underset{n\to +\infty}{\longrightarrow}0\textrm{, but in addition, we have that }\sum\limits_{n\geq 1}{\frac{1}{\left(n-1\right)!}}\textrm{ converges}\right) $ 
