I am asked to compute the stationary distribution of the markov chain with state space $E=\{0\dots,n\}$ and transition matrix below:
\begin{bmatrix} 0 & 1 \\ \frac{1}{n} & 0 & \frac{n-1}{n} \\ & \frac{2}{n} & 0 & \frac{n-2}{n} & \\ & & \ddots & \ddots & \ddots \\ & & & \frac{n-1}{n} & 0 & \frac{1}{n} \\ & & & & 1 & 0 & \\ \end{bmatrix}
I used $\pi P =\pi$ and I got:
$\pi_0=\frac{1}{n}\pi_1 \\\pi_1 = \pi_0+\frac{2}{n}\pi_2 \\ \pi_2 =\frac{n-1}{n}\pi_1 +\frac{3}{n}\pi_3 \\ \pi_3 =\frac{n-2}{n}\pi_2 +\frac{4}{n}\pi_4\\ \pi_4 =\frac{n-3}{n}\pi_3 +\frac{5}{n}\pi_5\\ $
Working from there I got:
$\pi_0 = \frac{1}{n}\pi_1 \\\pi_1 =n.\pi_0 \\ \pi_2=\frac{n^2-n}{2} \pi_0 \\ \pi_3 = \frac{n^3-3n^2+2n}{6} \pi_0 \\ \pi_4 = \frac{n^4-6n^3+11n^2-6n}{24} \pi_0 \\ \sum_{k=0}^\infty \pi_k = 1$
I tried fiddling with it and spotting a pattern for $\pi_k$ but I cant see to find $\pi_k$ for all $k\in E=\{0,\dots,n\}$. How would I finish this problem?
This is a follow-up from my previous question Calculating stationary distribution of markov chain , Sasha kindly showed me a way to find the stationary distribution, and I understood that method, but I don't think i've come across that method before in my lecture. I was wondering if it was possible to compute the stationary distribution using the method I did above, but I can't seem to get to the end..