$$ x_{n}=\sum^{n-1}_{i=0} {a_i x_{n-1-i}} $$ where $$ \sum^{+\infty}_{i=0} {a_i}=1,1>a_i>0,1>x_i>0 $$

In fact, the specific problem (comes from probability theory) I want to solve is that:

$0<d<0.2$ is a constant. $E_n(p) \in C[0,1]$is a function of p, $E_0(p)=p$, and $E_n(p)$ can be defined by：

$$ E_n(p)=a_n(p)+\sum^{n-1}_{i=0} {a_{n-1-i}(p) E_i(p)} $$

$$ if p>1-5d, then\ a_n(p)=0, $$$$ otherwise,\ a_n(p)=l_n(p) $$ where \begin{eqnarray} l_n(p)= \begin{cases} p, & n=0 \cr \prod^{n-1}_{k=0}{(1-p-d k)} (p+d n), & 1\leq n\leq 4 \cr \prod^{4}_{k=0}{(1-p-d k)} (1-p-5d)^{n-5} (p+5d),& n \geq 5 \cr \end{cases} \end{eqnarray}

I am sure that there is a function $u \in C[0,1]$ s.t. \begin{eqnarray} \lim\limits_{n\to+ \infty}|| E_n(p)-u(p)||_{max}=0 \end{eqnarray} but I do not know how to prove the convergence.

It is bounded, but it seems that the sequence is not monotonous.I tried some elementary methods to simplify the question, but we still need to prove $E_{n+1}-E_{n}$->0.

I found that $E_{n+1}-E_{n}$ is not decreasing, and I have no good ideas now.

By the way, I guess there is no easy(elementary) expression, but I cannot prove the existence yet.