I am asked to compute the stationary distribution of the markov chain with state space $E=\mathbb{N}_0$, $q_n >0$ for all $n \in \mathbb{N}_0$ and transition matrix below:
\begin{bmatrix} q_0 & q_1&q_2&q_3 &q_4&q_5&\dots \\ 1 & 0 & \\ & 1 & 0 & & \\ & & 1 & 0 & \\ & & & 1 & 0 & \\ & & & & 1 & 0 \\ & & & & & \ddots &\ddots \\ \end{bmatrix}
I used $\pi P = \pi$ and I also tried the way from my other thread Calculating stationary distribution of markov chain. Letting $g(x) = \sum_{k=0}^n x^k \pi_k$ What I got was:
$\pi_0 = \pi_1 +q_0\pi_0 \\ \pi_1 = \pi_2 +q_1\pi_0 \\ \pi_2 = \pi_3 +q_2\pi_0 \\ \pi_k =\pi_{k+1}+q_k\pi_0 \\ x^k\pi_k =x^k\pi_{k+1}+x^kq_k\pi_0 \\ \sum_{k=0}^{n}x^k\pi_k = \sum_{k=0}^{n}x^{k-1}\pi_{k}+\sum_{k=0}^{n}x^kq_k\pi_0 \\ g(x) =x^{-1}g(x)+\sum_{k=0}^{n}x^kq_k\pi_0$
and also
$\pi_0=\frac{1}{1-q_0}\pi_1 \\ \pi_1 =\pi_0(1-q_0) \\ \pi_2 =\pi_0(1-q_0-q_1) \\ \pi_3 =\pi_0(1-q_0-q_1-q_2) \\\pi_n = \pi_0(1-\sum_{k=0}^{n-1}q_k) \\$
I tried fiddling with it here and there but I cant seem to get anywhere to finish this problem. i.e. I can't seem to find $\pi_k$ for all $k \in E=\{0,\dots,n\}$. How would I finish this problem?
Edit: $\pi_n = \pi_0(1-\sum_{k=0}^{n-1}q_k) \\ \sum_{n=0}^{\infty}\pi_n =1 \\ \sum_{n=0}^{\infty}\pi_0(1-\sum_{k=0}^{n-1}q_k) =1 \\ \pi_0 = \frac{1}{\sum_{n=0}^{\infty}(1-\sum_{k=0}^{n-1}q_k)} = \frac{1}{(1-q_0)+(1-q_o-q_1)+(1-q_0-q_1-q_2)+\dots}$
Looking at @BrianMScott 's comment, how do I get the bottom line to simplify to $\sum_{k\ge 0}(k+1)q_k$? /To simplify the last line above?