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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.

Frederic Thomas
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    Possible duplicate of How can we use geometric series in calculating Green's function sum over loop effects – AccidentalFourierTransform Jan 06 '17 at 23:22
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    Hi Frederic, please see the link above: it seems to me that it is an exact duplicate of your question. I answered the other post; if you find it unconvincing, just leave a comment there and I'll try to add more details or to clarify some point that may remain unclear. – AccidentalFourierTransform Jan 06 '17 at 23:45
  • @AccidentalFourierTransform: Your post has convinced me. Just one question: Is there a simple reason why only the zeroth and first derivative of $\Sigma(p)|\bar{p}=m$ are divergent (without consideration of the counterterms)? – Frederic Thomas Jan 09 '17 at 22:50
  • Hi Frederic. Yes, there is a reason, and it is called renormalisability. QED is renormalisable, and you can find the proof in e.g. Weinbergs book. If $\Sigma''(\not p)$ were divergent, it would be impossible to absorb the divergence into a counterterm, and the theory would no longer be renormalisable. As the only counterterms you have are $\delta_m +\delta_1\not p$, the only divergences can be of the form $A+B\not p$, or otherwise they cannot be cancelled by the $\delta_i$ terms. Therefore, QED is renormalisable if $\Sigma(\not p)=A+B\not p+\text{finite}$. – AccidentalFourierTransform Jan 09 '17 at 23:01
  • @AccidentalFourierTransform: Okay. By the way, how do you do the Feynman-slash / in Latex ? – Frederic Thomas Jan 09 '17 at 23:07
  • I use the command \not, as in \not p (this produces $\not p$). For future reference, if you see some mathematical formula and want to know the commands that were used, you can right-click on in and select "Show Math As > TeX Commands". This will open up a window with the latex commands. Cheers! – AccidentalFourierTransform Jan 09 '17 at 23:10
  • @AccidentalFourierTransform : Thank very much for all the information. – Frederic Thomas Jan 09 '17 at 23:49

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