Gain of Inverting Active Bandpass Filter
Here is one way I can justify the approximation to be \$\frac{R_2}{2 R_1}\$hold only for Narrowband Filters. For bandpass action we would need \$\omega_1:=(R_1 C_1)^{-1} &lt; \omega_2:=(R_2 C_2)^{-1}\$. Now this allows us to write the gain as $$A=\frac{R_2}{R_1} \frac{\omega_1^{-1}}{\omega_2^{-1}+\omega_1^{-1}}.$$
This can also be expressed as $$A=\frac{R_2}{R_1}\frac{1}{1+\frac{\omega_1}{\omega_2}}.$$ Assuming the ratio \$\frac{\omega_1}{\omega_2} \to 1 \$ we get \$A \to R_2/(2 R_1)\$. This ratio approaching 1 signifies Narrowband Filters. 

This answer is summarizing the discussion with Dan Boschen, for Wideband filters (\$\frac{\omega_1}{\omega_2}\to 0\$), we get the approximation of the gain to be \$\frac{R_2}{R_1}\$.