Similar questions have been already asked, but I admit, by reading the answers I haven't got a satifying answer so far.
Let be $\bar{p}=\gamma^{\mu} p_{\mu}$ (Sorry I didn't find the slashed letter).
So if I calculate the two-point function in an interacting QFT for the Dirac-electron self energy I get according to Peskin-Schroeder (7.23) the following:
$\frac{i}{\bar{p}-m_0 } + \frac{i}{\bar{p}-m_0 }\left(\frac{\Sigma(p)}{\bar{p}-m_0}\right)+ \frac{i}{\bar{p}-m_0 }\left(\frac{\Sigma(p)}{\bar{p}-m_0}\right)^2 + \ldots $
I observe that although I might assume that $\Sigma(p) \sim \alpha $ and $\Sigma(p)$ might contain counter terms which cancel out the cut-off dependent part, that if $\gamma^{\mu} p_{\mu} \rightarrow m_0$ the term $\left(\frac{\Sigma(p)}{\bar{p}-m_0}\right)$ is not at all small, it can be easily larger 1. Therefore the given series should NOT converge, i.e. not be equal to $\frac{1}{\gamma^{\mu} p_{\mu}-m_0 -\Sigma(p)}$.
P&S even mention that there are multiple poles in the series, but it seems the $\Sigma(p)$ cancels all of them and leads to a finite $\frac{1}{\gamma^{\mu} p_{\mu}-m_0 -\Sigma(p)}$. But why ?
The summation is only allowed for $\bar{p}$ far away from $m_0$ ? Apparently not, as in the next section of the book $\bar{p}\sim m_0$ is assumed.
I would really appreciate if somebody could give me a sound explanation. Thank you very much.
