How to prove the convergence of this kind of sequence?

Question

$$ x_{n}=\sum^{n-1}_{i=0} {a_i x_{n-1-i}} $$ 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.

Answer 1

If $p>1-5d$, then $a_n(p)=0$ for all $n\ge1$ and hence $E_n(p)=0$ for all $n\ge1$.

So, without loss of generality $0\le p\le1-5d$. Then, letting $E_{-1}:=1$, for all $n\ge0$ we get \begin{equation*} E_n=\sum_{i=-1}^{n-1}l_{n-1-i}E_i, \end{equation*} where $E_n:=E_n(p)$ and $l_n:=l_n(p)$.

Here is the key point: Using the fact that $(l_5,l_6,\dots)$ is a geometric progression, one can check that \begin{equation*} E_{n+1}=\sum_{j=0}^5 b_j E_{n-j} \tag{1} \end{equation*} for all $n\ge4$, where \begin{equation*} b_j:=l_j+(p+5d-1)l_{j-1}. \end{equation*} Moreover, one can check that \begin{equation} \text{$b_0,\dots,b_5$ are $>0$ and $\sum_{j=0}^5 b_j=1$. }\tag{2} \end{equation}

The general solution of the linear difference equation (1) is given by \begin{equation*} E_n=\sum_{k=0}^5 C_k(n) z_k^n, \tag{3} \end{equation*} where the $z_k$'s are the roots of the characteristic polynomial \begin{equation*} P(z):=z^6-\sum_{j=0}^5 b_j z^{5-j}=z^6-\sum_{j=0}^5 b_{5-j}z^j \end{equation*} and the $C_k(n)$'s are polynomials in $n$ (whose degrees are nonzero if the corresponding roots $z_k$ are multiple ones). Moreover, in view of (2), we have the following:

(i) $1$ is a root of $P(z)$; say, $z_0=1$.

(ii) The multiplicity of $z_0=1$ is $1$, since $P'(1)=6-\sum_{j=0}^5 b_j(5-j)\ge6-5\sum_{j=0}^5 b_j=1>0$. So, $C_0:=C_0(n)$ does not depend on $n$.

(iii) $|z_k|<1$ for all $k=1,\dots,5$: Indeed, if $0=P(z)=z^6-\sum_{j=0}^5 b_{5-j}z^j$ for some complex $z$ with $|z|\ge1$, then, by the triangle inequality, $|z|^6\le\sum_{j=0}^5 b_{5-j}|z|^j\le\sum_{j=0}^5 b_{5-j}|z|^5=|z|^5$, so that $|z|\le1$ and hence $|z|=1$. Moreover, similarly the condition $|z|=1$ for a root $z$ of $P(z)$ implies the contradictory inequality $|z|^6<|z|^5$ unless the complex numbers $1$ and $z$ lie on the same ray emanating from $0$ -- that is, unless $z=1$.

It follows from (3) and (i)--(iii) that \begin{equation} E_n\to C_0 \tag{4} \end{equation} as $n\to\infty$. Of course, $E_n$ and $C_0$ may/will depend on $d,p$. The convergence in (4) holds for each pair $(d,p)$ satisfying the OP conditions $0<d<1/5$ and $0\le p\le1-5d$.