Virial expansion in terms of pressure I'm studying thermodynamics and I found two forms for the virial expansion:
$$pV~=~RT[1+\frac{A_2}{V}+\frac{A_3}{V^2}+\ldots] \tag{1}$$ and
$$pV~=~RT[1+B_2p+B_3p^2+\ldots]\tag{2}$$ my problem is that I can not find the correct procedure to  express the coefficients $B_k$ in terms of $A_k$. I just find the answer, that is $$B_2=\frac{A_2}{RT}, \qquad B_3=\frac{A_3-(A_2)^2}{(RT)^2}, \qquad \ldots \tag{3}$$ but I can not find the procedure to obtain those relations (Actually I found a very strange procedure that I didn't understand at all) and I have been trying but I can't solve this problem. Does anybody could help me please?.  
 A: Hints: 


*

*Define a "density/inverse volume" 
$$\alpha~:=~\frac{RT}{V}, \tag{A}$$
and rescale the $T$-dependent virial coefficient functions
$$ A_k^{\prime}~:=~\frac{A_k}{(RT)^{k-1}} . \tag{B}$$

*Then OP's 2 virial expansions (1) & (2) read
$$p~\stackrel{(1)}{=}~\alpha[1+A^{\prime}_2\alpha+A^{\prime}_3\alpha^2+\ldots] \tag{1'}$$ and
$$\alpha~\stackrel{(2)}{=}~p[1+B_2p+B_3p^2+\ldots]^{-1},\tag{2'}$$
respectively. 

*Next, in eq. (2') use the formulas for reciprocal power series:
$$\alpha~\stackrel{(2')}{=}~p[1+B^{\prime}_2p+B^{\prime}_3p^2+\ldots].\tag{2''}$$

*Finally eqs. (1') & (2'')  are mutually connected via power series reversion.
A: $$pV=RT\left [1+\frac{A_2}{V}+\frac{A_3}{V^2}+...\right ]\\\Rightarrow p=RT\left [\frac 1V+\frac{A_2}{V^2}+\frac{A_3}{V^3}+...\right] \\\Rightarrow p^2 = R^2T^2\left [ \frac {1 }{V^2}+ \frac{2A_2}{V^3}+...\right]$$
Now substitute for $p$ and $p^2$ into the equation  
$$pV=RT[1+B_2p+B_3p^2+...]$$  
and gather up terms is $\dfrac 1V$ and $\dfrac{1}{V^2}$ and then compare them with the first equation.
