This question is related to this question Proof of geometric series two-point function.
Suppose we have a graph $A$ with a symmetry factor $s_1$. According to Srednicki (chapter 9, eq. (9.13)) for a graphic $A^n$ we have an addition symmetry factor $n!$ so that the total symmetry factor of the graph $A^n$ is
$$s_n=n!s_1^n.$$
Now to proof that $$G_c^{(2)}(x_1,x_2)=[p^2-m^2+\Pi(p)]^{-1}$$ whe have to show that
$$G_c^{(2)}(x_1,x_2)=G_0 + G_0\Pi G_0+G_0\Pi G_0\Pi G_0+G_0\Pi G_0\Pi G_0\Pi G_0+...$$
$\Pi$ is the some of all irreducible graph so $\Pi$ is given by
$$\Pi= A+B+C+...$$
with $A,B,C,...$ irreducible graphs.
Now if we focus only on the graph $A$ we would have (including the symmetry factors)
$$G_c^{(2)}(x_1,x_2)=G_0 + G_0(\frac{AG_0}{s_1})+\frac{1}{2!}G_0(\frac{AG_0}{s_1})^2+\frac{1}{3!}G_0(\frac{AG_0}{s_1})^3+...$$ which is a not a geometric series.
What am I doing wrong here?
