TITLE: How to find the limit of this recurrence relation
QUESTION [11 upvotes]: I am trying to solve a question which requires me to solve this problem.

Given that $a_0=a_1=a_2=a_3=a_4=0$ and $a_5=1$ and
$$a_{n+6}=\dfrac{a_{n+5}+a_{n+4}+a_{n+3}+a_{n+2}+a_{n+1}+a_{n}}{6}$$ find the limit of$a_n$ as $n$ approaches infinity.

I tried various approaches but they do not seem to work, such as finding the fixed point of recurrence or finding the limit from the closed form of recurrence which do not exist. However, I calculated the values of $a_n$ up to $10000$ and the answer seems to extremely close to $2/7$
I am not looking for complete solution, I would rather appreciate if someone could point me in the right direction, just a hint would suffice.

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Following $\texttt{@Robert Z}$$\,\,\,$ answer:

\begin{align}
\mc{F}\pars{z} & \equiv
{6z^{5} \over 6 - z - z^{2} - z^{3} - z^{4} - z^{5} - z^{6}} =
{6z^{5} \over 6 - z\pars{1 - z^{6}}/\pars{1 - z}}
\\[5mm] & =
6\,{z^{5} - z^{6} \over z^{7} - 7z + 6} =
\sum_{p}r_{p}\pars{{1 \over z - p} + {1 \over p}}
\\[5mm] \mbox{where}\qquad &
\left\{\begin{array}{l}
\ds{p^{7} -7p + 6 = 0}
\\[2mm]
\ds{\left.r_{p}\right\vert_{\ p\ \not=\ 1} \equiv
{6 \over 7}\,{p^{5}\pars{1 - p} \over p^{6} - 1}\,,\qquad r_{1} \equiv
-\,{2 \over 7}}
\\[2mm]
\ds{r_{p}}\ \mbox{is the}\ residue\ \mbox{at pole}\ p.
\end{array}\right.
\\[5mm] 
\mbox{Note that}\quad &
\left.r_{p}\right\vert_{\ p\ \not=\ 1} = {1 \over 7}\,{p^{6}\pars{1 - p} \over p^{7} - p} =
{6 \over 7}\,{p^{6}\pars{1 - p} \over \pars{7p - 6} - p} =
{1 \over 7}\,{p^{6}\pars{1 - p} \over p - 1} = -\,{p^{6} \over 7}
\end{align}


With $\ds{0 < a < \min\braces{\verts{p}}}$:

\begin{align}
a_{n} & =
\oint_{\verts{z} = a}{\mc{F}\pars{z} \over z^{n + 1}}\,{\dd z \over 2\pi\ic} =
{1 \over n!}\,\lim_{z \to 0}\,\totald[n]{\mc{F}\pars{z}}{z} =
{1 \over n!}\,\lim_{z \to 0}
{\sum_{p}r_{p}\,{\pars{-1}^{n}n! \over \pars{z - p}^{n + 1}}} =
-\sum_{p}{r_{p} \over p^{n + 1}}
\\[5mm] & = - r_{1} + {1 \over 7}\sum_{p \not= 1}{1 \over p^{n - 5}} =
{2 \over 7} + {1 \over 7}\sum_{p \not= 1}{1 \over p^{n - 5}}
\implies \bbx{\lim_{n \to \infty}a_{n} = {2 \over 7}}
\end{align}

The $\ds{\,\mc{F}\pars{z}}$ poles, which are different of one, have magnitude greater than $\ds{\color{#f00}{one}}$ !!!.