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