Stationary distribution of an infinite Markov chain The exercise is asking for the stationary distribution, the estimated time to get from state $0$ to state $4,$ and to conclude if the chain is time-reversible.
So I have the following transition matrix:
$$P=\left[\begin{matrix}
1/2&1/2&0&0&0&\cdots\\
1/2&0&1/2&0&0&\cdots\\
1/2&0&0&1/2&0&\cdots\\
1/2&0&0&0&1/2&\cdots\\
1/2&0&0&0&0&\cdots\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots
\end{matrix}\right]$$
This means $P_{i,0} =1/2,$ $P_{i,i+1} = 1/2,$  and $P_{ij}=0$ in every other case.
I have progressed up to this point:
\begin{align*}
n_1&=\frac12\,n_1+\frac12\,n_2+\frac12\,n_3+\dots\\
n_2&=\frac12\,n_1\\
n_3&=\frac12\,n_2\\
&\vdots\\
n_k&=\frac12\,n_{k-1}\\
\sum_{k=1}^\infty n_k&=1.
\end{align*}
I'm not sure how to solve the problem, though.
Any help would be very much appreciated.

We can rewrite manipulate the first equation by noticing that $\pi_1+\pi_2+\pi_3+\cdots=1$. This gives us the following:
$$\begin{align}
\pi_1&=\frac12\pi_2+\frac12\pi_3+\frac12\pi_4+\cdots\\
&=\frac12\left(\pi_2+\pi_3+\pi_4+\cdots\right)\\
&=\frac12(1-\pi_1)
\end{align}$$
Solving $\pi_1=\frac12(1-\pi_1)$ gives us $\pi_1=\frac13$. Then, each term is half of the previous one ($\pi_n=\frac12\pi_{n-1}$), so we have $\pi_n=\frac13\left(\frac12\right)^{n-1}$.
Hope it was helpful!
