Sitting on the fence

Simple math with the values give:

$g(n) = g(n-1) - 4\cdot g(n-2)$

Also, calculating $g(-1)$ and $g(-2)$ from $g(1)$ and $g(2)$ gives $g(-1) = \frac{-1}{4}$ and $g(-2) = \frac{-1}{16}$

Applying the Z-transform on $g(n) = g(n-1) - 4\cdot g(n-2)$ and using the values of $g(-1)$ and $g(-2)$ previously calculated, gives:

$G(z) = \frac{z^{-1}}{1 - z^{-1} + 4z^{-2}}$

$G(z) = \frac{k_1}{1 - \frac{1 + i\sqrt(15)}{2} \cdot z^{-1}} + \frac{k_1}{1 - \frac{1 - i\sqrt(15)}{2} \cdot z^{-1}}$

Which will lead to:

$g(n) = k_1 \cdot (\frac{1 + i\sqrt{15}}{2})^n + k_2 \cdot (\frac{1 - i\sqrt{15}}{2})^n$

So, each term of the series s will be in the form of $g(n)$ divided by $2^n$ , or:

$s(n) = k_1 \cdot (\frac{1 + i\sqrt{15}}{4})^n + k_2 \cdot (\frac{1 - i\sqrt{15}}{4})^n$

Thus, s will be a sum of two geometric progressions, one of ratio $\frac{1 + i\sqrt{15}}{4}$ and the other of ratio $\frac{1 - i\sqrt{15}}{4}$ . Since $\mid \frac{1 + i\sqrt{15}}{4} \mid= \mid\frac{1 - i\sqrt{15}}{4} \mid= 1$ , the series diverge, and thus, the answer is $\fbox{0.666}$

The numerators are given by $g(n)=g(n-1)-4g(n-2)$ , and the terms of the series are $\frac{1}{2}U_{n-1}(\frac{1}{4})$ $=\frac{2\sin(n\arccos(\frac{1}{4}))}{\sqrt{15}}$ , where $U_n(x)$ is the Chebyshev polynomial of the second kind . Since these terms fail to converge to $0$ , the series diverges. The answer is $\boxed{0.666}$