IRL ABooklet

Suppose, for some $r > 0$, a two-way infinite sequence $a(n)$ satisfies
$$a(-n) = 1 - \sum_{k=0}^{n} r^k a(n-k), \quad \text{for all } n \in \mathbb{Z}. \qquad (1)$$

For $n = 0$ this means $a(0) = 1 - a(0)$, hence $a(0) = \frac{1}{2}$. For $n = \pm 1$ we get
$$a(1) + a(-1) = 1 - r^{-1}a(0) \quad \text{and} \qquad (2)$$
$$a(-1) + a(1) = 1 - ra(0). \qquad (3)$$
Combining (2) and (3) gives $r^{-1}a(0) = ra(0)$, whence $r^2 = 1$, so that $r = 1$, as claimed. With $r = 1$, (1) simplifies to

$$a(-n) = 1 - \sum_{k=0}^{n} a(n-k), \quad \text{for all } n \in \mathbb{Z}.$$
We rewrite this in two ways, using only $n \ge 0$:
$$a(-n) = 1 - \sum_{k=0}^{n} a(k) \quad \text{and} \qquad (4)$$
$$a(n) = 1 - \sum_{k=0}^{n} a(-k). \qquad (5)$$

Writing $d(n) = a(n) - a(-n)$, the difference of (4) and (5) gives
$$d(n) = \sum_{k=0}^{n} d(k), \text{ i.e. } \sum_{k=0}^{n-1} d(k) = 0, \quad \text{for all } n \ge 1, \qquad (6)$$
Clearly, $d(0) = a(0) - a(-0) = 0$ and so induction and (6) gives $d(n) = 0$, i.e. $a(n) = a(-n)$ for all $n \ge 0$. Now (5) becomes
$$a(n) = 1 - \sum_{k=0}^{n} a(k), \quad \text{i.e.}\quad 2a(n) = 1 - \sum_{k=0}^{n-1} a(k), \quad \text{for all } n \ge 0.$$
Replacing $1 - \sum_{k=0}^{n-2} a(k)$ in this last expression by $2a(n-1)$, we obtain
$$2a(n) = 2a(n-1) - a(n-1) = a(n-1), \quad \text{for all } n \ge 1.$$
Since $a(0) = 1/2$ it now follows by induction that $a(n) = 1/2^{n+1}$, for all $n \ge 0$. Because $a(-n) = a(n)$, we finally obtain
$$$$
a(n) = $\frac{1}{2^{|n|+1}}, \quad \text{for}$ all } n $\in \mathbb{Z}.$