jmc

Note that $\lfloor {\tau }^0\rceil =\lfloor 1\rceil =1$ and $\lfloor \tau \rceil =2.$

Let $\sigma =\frac{1-\sqrt{5}}{2},$ and let $L_n={\tau }^n+{\sigma }^n.$ Then $$\begin{array}{rl}{L}_n& ={\tau }^n+{\sigma }^n\\ & =(\tau +\sigma )({\tau }^{n-1}+{\sigma }^{n-1})-\tau \sigma ({\tau }^{n-2}+{\sigma }^{n-2})\\ & =L_{n-1}+L_{n-2}.\end{array}$$Also, $L_0=2$ and $L_2=1,$ so $L_n$ is an integer for all $n\ge 0.$

Furthermore, $${\sigma }^2=\frac{3-\sqrt{5}}{2}<\frac{1}{2},$$so for $n\ge 2,$ $|{\sigma }^n|<\frac{1}{2}.$ Hence, $$\lfloor {\tau }^n\rceil =L_n$$for all $n\ge 2.$

Let $$S=\frac{L_2}{2^2}+\frac{L_3}{2^3}+\frac{L_4}{2^4}+\cdots .$$Then $$\begin{array}{rl}S& =\frac{L_2}{2^2}+\frac{L_3}{2^3}+\frac{L_4}{2^4}+\cdots \\ & =\frac{L_0+L_1}{2^2}+\frac{L_1+L_2}{2^3}+\frac{L_2+L_3}{2^4}+\cdots \\ & =\left(\frac{L_0}{2^2}+\frac{L_1}{2^3}+\frac{L_2}{2^4}+\cdots \right)+\left(\frac{L_1}{2^2}+\frac{L_2}{2^3}+\frac{L_3}{2^4}+\cdots \right)\\ & =\left(\frac{1}{2}+\frac{1}{8}+\frac{S}{4}\right)+\left(\frac{1}{4}+\frac{S}{2}\right).\end{array}$$Solving, we find $S=\frac{7}{2}.$

Therefore, $$\sum _{n=0}^{\infty }\frac{\lfloor {\tau }^n\rceil }{2^n}=1+\frac{2}{2}+\frac{7}{2}=\boxed{\frac{11}{2}}.$$