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

Question

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

$$\begin{align} D'_{\mu\nu}(k)&=\frac{1}{k^2[1+\Pi(k^2)]}\Bigg(-g_{\mu\nu}-\frac{k_\mu k_\nu}{k^2}\Pi(k^2)\Bigg)\\&=\frac{-g_{\mu\nu}}{k^2[1+\Pi(k^2)]} + \text{gauge terms}. \end{align}\tag{9.124}$$

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

Answer 1

OP asks a very good conceptional question that goes to the heart of renormalization. Consider Ryder's formula for the self-energy

$$\begin{align} \Pi~=~&\frac{e^2}{6\pi^2}\left(\frac{1}{\epsilon}+\frac{k^2}{10m^2} \right)+{\cal O}(e^4)\cr ~=~&\frac{e^2}{6\pi^2\epsilon} + \Pi_{\rm finite}+{\cal O}(e^4).\end{align}\tag{9.123} $$

The answer to OP's question is that in the renormalization procedure we treat the self-energy$^1$ $$\Pi~=~ \sum_{n=1}^{\infty} (e^2)^n\Pi_n \tag{A}$$ as a formal power series [without a constant term] in the coupling constant $e^2$. Each coefficient $$\Pi_n=\sum_{m=-N}^{\infty}\epsilon^m \Pi_{nm}\tag{B}$$ is a truncated Laurent series in $\epsilon$. The coefficients $\Pi_n$ are not necessarily small, as OP already has observed. For this and other reasons, the power series (A) is not convergent. However, it still makes sense to treat it as a formal power series. Formal power series form an algebra with a well-defined addition and multiplication. In this way we are able to make consistent perturbative calculations even if the coefficients $\Pi_n$ are not small.

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$^1$ More precisely, the underlying $Z$-factors.