Charge Renormalization in Peskin and Schroeder I'm working on the charge renormalization in Peskin(page 252).

However I don't understand how he got infinite bare charge $e_0$.
According to $$\frac{e^{2}-e_{0}^{2}}{e_{0}^{2}}\approx-\frac{2 \alpha}{3 \pi \epsilon}$$
Solve this equation and we get $$e_0^2=\frac{3 \pi\epsilon}{3 \pi \epsilon-2 \alpha}e^2$$
When $\epsilon \rightarrow 0$, we have
$$e_0^2=-\frac{3 \pi \epsilon}{2 \alpha}e^2+O\left(\epsilon^{2}\right)$$
So the bare charge $e_0^2$ should be infinitely smaller than $e^2$.
What's wrong with my derivation？
 A: You should remember that $\alpha = e^2/4\pi$, and that your expressions should all be determined perturbatively in $e$. Since you have only computed corrections to your theory to order $e^4$, the solution to the equation should really be written
$$
e_0^2 = \frac{3 \pi \epsilon}{3 \pi \epsilon - e^2/2\pi} e^2 \approx e^2 + \frac{e^4}{6 \pi^2 \epsilon}.
$$
After all, by keeping the $e^2$ term in the denominator of your equations, you are implicitly keeping higher-order terms which will receive further corrections from higher loops.
In other words, one should take the $e \rightarrow 0$ limit before the $\epsilon \rightarrow 0$ limit. The reasons for this come from some of the mathematical details behind dimensional regularization, which relies on the fact that the renormalization functions like $Z_3$ take the form
$$
Z_3 = 1 + \sum_{n = 1}^{\infty} \frac{a_n(e^2)}{\epsilon^n} + \ \mathrm{regular \ terms \ in} \ \epsilon,
$$
with coefficients that have coupling dependence like $a_n(e^2) = O(e^{2(n+1)})$. 
