I have a markov chain with transition matrix below,
$\begin{bmatrix} 1-q & q & & & \\ 1-q & 0 & q & & \\ & 1-q & 0 & q & \\ & & 1-q & 0 & q \\ & & & \ddots & \ddots & \ddots \end{bmatrix}$ and I am asked to compute the stationary distribution for $q<\frac{1}{2}$. Using $\pi P =\pi$, I get that $\pi_n =\pi_0\left(\frac{q}{1-q}\right)^n$ and $\sum_{n=0}^\infty \pi_n = 1$.
Thus I get
$\pi_0\sum_{n=0}^\infty \left(\frac{q}{1-q}\right)^n=1 \\ \pi_0=\frac{1}{\sum_{n=0}^\infty \left(\frac{q}{1-q}\right)^n} \\ \text{thus } \pi_n = \frac{\left(\frac{q}{1-q}\right)^n}{\sum_{n=0}^\infty \left(\frac{q}{1-q}\right)^n}$
But it doesnt seem to be simplified. Is there anything else I can do to simplify $\pi_0=\frac{1}{\sum_{n=0}^\infty \left(\frac{q}{1-q}\right)^n}$? And thus simplify $\pi_n$.