Combinatorics geometric series two-point function In this answer Proof of geometric series two-point function it is said: 


  
*Now what about the coefficients in front of each Feynman diagram? Due to the combinatorics/factorization involved it becomes a geometric series 
  $$G_c~=~G_0\sum_{n=0}^{\infty}(\Sigma G_0)^n\tag{A}$$

How can we prove this? My main concern is the combinatorics, for example in qed the symmetry factor for connected diagrams is 1.suppose that $\Sigma=A+B+...$ are irreducible diagrams of the photon propagator.
Since symmetry factor of $G_0AG_0AG_0=1$  we should have the symmetry factor of $A=1$.The same thing for $B$.
But we also have the factor $G_0AG_0BG_0+G_0BG_0AG_0$
The only way this to work is that $G_0AG_0BG_0=-G_0BG_0AG_0$ 
How can I prove this?
 A: *

*Recall

*

*that the partition function $Z[J]$ is the generating functional of all$^1$ Feynman diagrams;


*that $$W_c[J]~=~\frac{\hbar}{i}\ln Z[J]$$ is the generating functional of connected Feynman diagrams, cf. the linked cluster theorem;


*and that the Feynman rules dictate that each such Feynman diagram should be divided by its symmetry factor.




*In contrast, for a connected $n$-point correlation function $$\langle\phi^{i_1}\ldots\phi^{i_n}\rangle^c_J~=~ \left(\frac{\hbar}{i}\right)^{n-1}\frac{\delta^n W_c[J]}{\delta J_{i_1} \ldots\delta J_{i_n}},$$ the $n$ external legs are considered distinguishable, and not symmetrized. We emphasize that it does not contain a $S_n$-symmetry factor of its $n$ external legs.


*In particular,

*

*the full propagator/connected 2-pt function $$\langle\phi\phi\rangle^c_{J=0}~=~\frac{\hbar}{i}G_c$$ from eq. (A),


*the bare propagator $\frac{\hbar}{i}G_0$, and


*the self-energy $\frac{i}{\hbar}\Sigma$
are not divided by the $\mathbb{Z}_2$-symmetry of their 2 external legs, independently of whether the propagator is directed or undirected/has or hasn't an arrow.


*Similarly, if the self-energy $$\frac{i}{\hbar}\Sigma~=~\sum_k\frac{F_k}{S_k}$$ is built from individual 1PI Feynman diagrams $F_k$ with 2 (amputated) external legs, we only divide with symmetry $S_k$ of $F_k$ as if the 2 external legs were distinguishable. Furthermore, the geometric series (A) therefore neatly generates connected Feynman diagrams weighted with appropriate symmetry factors.
References:

*

*P. Etingof, Geometry & QFT, MIT 2002 online lecture notes; Chapter 3.

--
$^1$ For the perturbative expansion, see e.g. Ref. 1 or my Phys.SE answer here.
