Find the limit of $f_{n}(1)$ if $f_{n}(x)=\int^{x}_{0}f_{n-1}(y)\,dy$ for each $n$ 
Consider the sequence $(f_n)$ defined by
  $$
f_{0}(x)=\frac{1}{(1+x)^3}\quad
f_{n}(x)=\int^{x}_{0}f_{n-1}(y)\;\text{dy},\ n\ge1.$$
  Find 
  $$\lim_{n\to \infty}f_{n}(1).$$

I computed $f_{1}(x)$, $f_{2}(x)$, $f_{3}(x)$, but I couldn’t get it that way.
 A: Hint: Show by induction on $n$ that, for every $n\geqslant1$ and every $x$ in $[0,1]$, $$|f_n(x)|\leqslant\|f_0\|_\infty\,\frac{x^n}{n!}$$ and deduce that the desired limit is... 

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad0$

Comment: Being given a specific function $f_0$ can actually be a distraction.
A: Notice that, by convexity,
$$ \frac{1}{8}\leq f_0(x) \leq 1-\frac{7x}{8}\tag{1} $$
so, integrating the previous inequality multiple times, we get:
$$ \frac{x^n}{8n!}\leq f_n(x) \leq \frac{x^n}{n!}-\frac{7}{8}\frac{x^{n+1}}{(n+1)!} \tag{2} $$
from which it follows that:
$$ \frac{1}{8n!} \leq f_n(1) \leq \frac{8n+1}{8(n+1)!}. \tag{3} $$
A: If I'm not mistaking you can transform your integral into
$$f_n(x)=f^{(-n)}(x)=\int_{0}^x\int_{0}^{a_1}\int_0^{a_2}\cdots\int_0^{a_{n-1}}f_0(a_n)da_nda_{n-1}\cdots da_2da_1=\\=\frac{1}{(n-1)!}\int_{0}^{x} (x-t)^{n-1}\frac{1}{(1+t)^3}$$
by using the Cauchy formula for repeated integration,the integral converges for $x=1$ and $\frac{1}{(n-1)!}$ diverges (if I'm not mistaking)
