Is the induced drag independent of wing span?
Aspect ratio $AR$ can be written as $\frac{b}{c}$ which is equal to $\frac{b^2}{S}$. 

Before we start substituting, note that $C_L$ also depends on the wing surface area $S$.

$L = \frac{1}{2}\rho V^2 C_L S$

or

$C_L =  \frac{L }{\frac{1}{2} \rho V^2 S}$

Substituting all this in the induced drag formula yields:

$D_i = \frac{1}{2}\rho V^2 S \frac{C_L^2}{\pi  AR \epsilon} =
\frac{L^2}{\frac{1}{2}\rho V^2 S \pi AR \epsilon} = \frac{L^2}{\frac{1}{2}\rho V^2   \pi b^2 \epsilon}$

This shows that the induced drag is proportional to the inverse of the square of the wingspan.