Suppose that the chain is intitially in state $1$, i.e $P(X_0 = 1) = 1$. Let $\tau$ denote the time of first returen to state $1$, i.e
$$\tau = \min\{n > 0: X_N = 1\}.$$
Show that
$$P(\tau = k) = (0.5)^{k-1}, k = 2, 3, ...$$
State $1$ only communicates with state $4$. I have already (correctly) got that $P_{14} = 1, P_{44} = \frac{1}{2}, P_{41} = \frac{1}{2}$.
So to do this question, what basically happens is that my process will first go from $1$ to $4$. It will then stay in $4$ for some time $k$. Then, it will either remain in $4$ or go back to $1$. The probability of this happening is
$$P_{14} \times P_{44}^k \times P_{41} = 1 \times (0.5)^k \times (0.5) = (0.5)^{k+1}$$
which isn't the right answer.
Where have I gone wrong?
Also, the next part tells me that using this relation and the definition of recurrence, I need to verify that state $1$ is recurrent. In the answers, they say
We need to show that $P(\tau = \infty) = 1$. Observe that
$$P(\tau < \infty) = \sum_{k = 2}^{\infty} P(\tau = k) = \sum_{k = 2}^{\infty} (0.5)^{k-1} = \sum_{j = 1}^{\infty} (0.5)^j = \frac{0.5}{1 - 0.5} = 1$$
How have they managed to do this. I get what we want to show, due to the definition of recurrence, but why have they then worked it out for $\tau < \infty$ and how have they gone between each of the summation signs to get $\frac{0.5}{1-0.5}$?
