How can the connected Greens 2pt function be summed as a geometric series of the self-energy if the self-energy contains divergent terms?

In Ryder’s book on QFT page 341 we can see
$$begin{align}
D_{munu}’=D_{munu}-D_{mualpha}big(k^alpha k^beta-g^{alphabeta}k^2big)Pi(k^2)D_{betanu}
end{align}$$
and hence putting $D_{munu}=-g_{munu}/k^2,$
$$begin{align}
D’_{munu}(k)&=frac{1}{k^2[1+Pi(k^2)]}Bigg(-g_{munu}-frac{k_mu k_nu}{k^2}Pi(k^2)Bigg)&=frac{-g_{munu}}{k^2[1+Pi(k^2)]} + text{gauge terms}.
end{align}tag{9.122}$$
I don’t understand how he derived this equation,
I tried to derive the last expression as follows
putting $D_{munu}=-frac{g_{munu}}{k^2}$(Feynman’s propagator) gives
$$begin{align}
D’_{munu}(k)=frac{-g_{munu}}{k^2}-Bigg(frac{k_mu k^nu}{k^2}-g_{munu}Bigg)frac{Pi(k^2)}{k^2}
=frac{1}{k^2}big(1-Pi(k^2)big)Bigg[-g_{munu}-frac{k_mu k^nuPi(k^2)}{k^2(1-Pi(k^2))}Bigg].end{align}$$
Iff $Pi(k^2)ll 1$ we can use the expansion
$$(1-x)^{-1}=1+x+x^2+x^3+x^4ldots$$
On using this expansion we get

$$begin{align}
D’_{munu}(k)&=frac{1}{k^2[1+Pi(k^2)]}Bigg(-g_{munu}-frac{k_mu k_nu}{k^2}Pi(k^2)Bigg)&=frac{-g_{munu}}{k^2[1+Pi(k^2)]} + text{gauge terms}.
end{align}tag{9.124}$$

But in fact, $Pi(k^2)gg 1$ due to the presence of the divergent term $frac{1}{6pi^2epsilon}$ in $Pi(k^2)$. How can we say Ryder is correct?