Proof for the expression of Boyle's temperature

If you recall the Van der Waals equation, $$left(p+frac{a}{V_m}right)(V_m-b)=RT$$ where $V_m=V/n_text{moles}$ is the molar volume. $$p=frac{RT}{V-b}-frac{a}{V^2}=frac{RT}{V}left(1-frac{b}{V}right)^{-1}-frac{a}{V^2}$$ and using the binomial expansion, the terms in brackets can be expanded into a series, resulting in $$frac{pV}{RT}=1+frac{1}{V}left(b-frac{a}{RT}right)+left(frac{b}{V}right)^2+left(frac{b}{V}right)^3+cdots$$ The Boyle temperature $T_B$ is defined by $B(T_B)=0$ and hence $$T_B=frac{a}{Rb}Rightarrow T_B=frac{27T_c}{8} $$ Near Boyle's temperature $$frac{pV}{RT}approx 1+mathcal{O}left(left(frac{b}{V}right)^2right)approx 1$$ So real gas approximately behave like ideal gases.