How to find the limit of this recurrence relation

Question

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.

Answer 1

I'm going to use the same idea as in this question:

Limit of sequence in which each term is defined by the average of preceding two terms

$$6a_{n+6}=a_{n+5}+a_{n+4}+a_{n+3}+a_{n+2}+a_{n+1}+a_{n}$$

Now calculate some cases:

$$6a_{6}=a_{5}+a_{4}+a_{3}+a_{2}+a_{1}+a_{0}$$ $$6a_{7}=a_{6}+a_{5}+a_{4}+a_{3}+a_{2}+a_{1}$$ $$6a_{8}=a_{7}+a_{6}+a_{5}+a_{4}+a_{3}+a_{2}$$ $$6a_{9}=a_{8}+a_{7}+a_{6}+a_{5}+a_{4}+a_{3}$$ $$6a_{10}=a_{9}+a_{8}+a_{7}+a_{6}+a_{5}+a_{4}$$ $$6a_{11}=a_{10}+a_{9}+a_{8}+a_{7}+a_{6}+a_{5}$$ $$...$$

when we keep writing and sum every equation we see that all terms $a_i$ for $6\le i\le n-6$ will be canceled on both sides.

We then will get:

$$6a_{n}+5a_{n-1}+4a_{n-2}+3a_{n-3}+2a_{n-4}+a_{n-5}=6a_{5}+5a_{4}+4a_{3}+3a_{2}+2a_{1}+a_{0}=6$$

and if $a_n\to L$ then

$$6L+5L+4L+3L+2L+L=6\to L=\frac{2}{7}$$

Answer 2

Hint: Compute the next few terms of the recurrence, which will give you a pretty clear clue as to what the closed form should be. You can prove this by induction, and then the limit should become clear.

Answer 3

The generating function of the sequence is $$f(z)=\sum_{n=0}^{\infty}a_nz^n=\frac{6z^5}{6-z-z^2-z^3-z^4-z^5-z^6}.$$ Note that one of the poles is $1$ and the others are all complex numbers outside the disc $|z|\leq 1$. Hence $$\lim_{n\to\infty}a_n=-\mbox{Res}(f,1)=\frac{6}{1+2+3+4+5+6}=\frac{2}{7}.$$ If we replace $6$ by $N$, by using the same approach, we find that the limit is $$\frac{N}{1+2+\dots+N}=\frac{2}{N+1}.$$

Answer 4

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

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}

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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 $\ds{\color{#f00}{one}}$ !!!.