Does this really have a closed form?

Let $\mathbf{S}$ denote $\displaystyle \sum_{n=0}^{\infty} \frac{F_{n}}{4^{n}}$ .Then we have $\begin{aligned} \mathbf{S} & =\displaystyle \sum_{n=0}^{\infty} \frac{F_{n}}{4^{n}}=\sum_{n=1}^{\infty} \frac{F_{n}}{4^{n}} \quad \quad[\because F_{0}=0] \\ & =\frac{1}{4}+\frac{1}{4^2}+\frac{2}{4^3}+\frac{3}{4^4}+\frac{5}{4^5}+\frac{8}{4^6} +\ldots\quad \quad & \ldots(1) \\ \implies \frac{\mathbf{S}}{4} & =\frac{1}{4^2}+\frac{1}{4^3}+\frac{2}{4^4}+\frac{3}{4^5}+\frac{5}{4^6}+\frac{8}{4^7}+\ldots \quad \quad & \ldots(2) \\ \implies \frac{3\mathbf{S}}{4} & =\frac{1}{4}+\frac{1}{4^3}+\frac{1}{4^4}+\frac{2}{4^5} +\frac{3}{4^6}+\frac{5}{4^7}+\ldots \quad[\small{\text{Subtracting}~ (2)~\text{from} ~(1)}] \\ \implies \frac{3\mathbf{S}}{4} & =\frac{1}{4}+\frac{\mathbf{S}}{16} \\ \implies \mathbf{S} & =\frac{4}{11} \\ \therefore 22\mathbf{S} & =\boxed{8} \end{aligned}$