Q: Why is perturbation theory always implemented around $\alpha=0$? In the perturbative approach to field theory we expand whatever we are computing on a power expansion in some coupling 
$$
\sum^nd_n\alpha^n
$$
then in principle we can compute all the $d_n$. This series is in general expected not to be convergent, but it is hoped that it at least is an asymptotic expansion of the true thing when $\alpha\to0$.
My question is, why is the perturbative approach only implemented around $\alpha=0$? I mean, we could make expansions around any given $\alpha_0$ and obtain expansions that (would be hoped) to be asymptotic to the real thing when $\alpha\to\alpha_0$?
$$
\sum^nb_n(\alpha-\alpha_0)^n
$$
why is perturbation theory always implemented around $\alpha=0$?

A: It is because one wants to do a perturbation theory around the known solution. Of course, you can denote by $\alpha$ anything you want, but usually it is some coupling constant. The limit $\alpha \rightarrow 0$ corresponds to a free theory, which can be easily solved. After it is done, one computes the corrections to the known solution to the free theory in the form of the perturbation series in $\alpha$.

A: Usually the problem is set such that $\alpha=0$ simplifies the equations: eliminates mixing / interacting terms, or allows to ignore certain effects at first order. In general, we want to get a problem that we can solve, that is the main point of perturbation theory. If we can solve the problem for some different value of the parameter, than it is meaningful to expand it around the other point.
Note that you can always rescale the parameter defining $\beta = \alpha - \alpha_0$ and have the new series in $\beta$ around at zero, so effectively the starting point, as far the parameter itself is concerned, is meaningless. However note that the coefficients $b_n$ and $d_n$ of the expansion around the two points are related, so they are effectively equivalent as long as you can actually solve the problem for both values of the parameter.
