How to find the limit of this recurrence relation

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 $$\frac{N}{1+2+\dots+N}=\frac{2}{N+1}.$$

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

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

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