Rational?

$\begin{aligned} (1-x)(1+x+x^2+x^3+x^4) & = \frac {31}{32} &...(*) \\ \implies 1-x^5 & = \frac {31}{32} \\ \implies x^5 & = 1 - \frac {31}{32} = \frac 1{32} \\ \implies x & = \frac 12 \\ \left(1-\frac 12 \right)(1+x+x^2+x^3+x^4) & = \frac {31}{32} &...(*) \\ \implies 1+x+x^2+x^3+x^4 & = \frac {31}{16} \\ \implies 1+x+x^2+x^3+x^4 + x^5 & = \frac {31}{16} + \frac 1{32} \\ & = \frac {63}{32} = \boxed{1.96875} \end{aligned}$

$\displaystyle \begin{aligned} (1 - x)(1 + x + x^2 + x^3 + x^4) & = \frac{31}{32} & \small \color{#3D99F6} \sum_{k \ = \ 0}^{n} ar^k = \frac{a(1 - r^{n + 1})}{1 - r} \\ (1 - x) \cdot \frac{1 - x^5}{1 - x} & = \frac{31}{32} \\ 1 - x^5 & = \frac{31}{32} \\ x^5 & = \frac{1}{32} \\ x & = \frac 12 \\ \\ \implies 1 + x + x^2 + x^3 + x^4 + x^5 & = \frac{1 - x^6}{1 - x} \\ & = \frac{1 - \frac{1}{64}}{1 - \frac 12} \\ & = 2 \cdot \frac{63}{64} \\ & = \boxed{\displaystyle \frac{63}{32}} \end{aligned}$

$(1-x)(1+x+x^{2}+x^{3}+x^{4}) = \dfrac{31}{32}$

$1+x+x^2+x^3+x^4-x-x^2-x^3-x^4-x^5=\dfrac{31}{32}$

$1-x^5=\dfrac{31}{32}$

$-x^5=\dfrac{31}{32}-1$

$x^5=\dfrac{1}{32}$

$x=\dfrac{1}{2}$

The answer is $\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+1=\boxed{1.96875}$