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

  • $\begingroup$ As for the question itself, we have $x_n=(1+a_{n-1})x_{n-1}$, so $\lim x_n=\prod (1+ a_n)$, and such an infinite product converges if $a_n$ is summable (take the logarithm to see this). $\endgroup$ Commented Jul 5, 2021 at 14:03
  • $\begingroup$ I think the question is rather nontrivial. So, I don't understand the close votes. $\endgroup$ Commented Jul 6, 2021 at 4:27
  • $\begingroup$ @ChristianRemling : I don't understand your comment. How do you get that recurrence? $\endgroup$ Commented Jul 6, 2021 at 4:31
  • $\begingroup$ @IosifPinelis: My comment refers to the original version of the question, before the edits. $\endgroup$ Commented Jul 6, 2021 at 14:39

1 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 $0\le p\le1-5d$.

  • $\begingroup$ Thank you so much.I just used this method to deal with the case of $p+5 d>1$ at first, because I forgot that infinite items can be converted into finite items. I finally solved this problem myself, but used a very troublesome method to estimate $E_{n+1}-E{n}$. Your formula (1) made me enlightened. $\endgroup$
    – zhjzwlys
    Commented Jul 6, 2021 at 5:38
  • $\begingroup$ +1 (I wonder why OP didn't upvote the answer) $\endgroup$
    – tj_
    Commented Jul 6, 2021 at 7:02

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