A proof I'm reading states without further explanation that
$$ -\frac{1}{2}\delta^2 = \frac{1}{2} \zeta^2 + \frac{\alpha}{6} \zeta^3 + \frac{\beta}{24} \zeta^4 +\cdots $$
can be inverted into
$$ \zeta = i\delta + \frac{\alpha}{6} \delta^2 +\Big[ \frac{\beta}{24} - \frac{5\alpha^2}{72} \Big] i\delta^3+\cdots, $$
where $\alpha$ and $\beta$ are some constants. How can this be done? Is there some formula giving the inverse expansion of such series?
At a first glance, it looks like it shouldn't be possible, because you have a singular case (no linear term), but fortunately, there is a square on the left, too, so you actually have a function
$$i\delta =\zeta\sqrt{1+\frac{\alpha}{3}\zeta+\frac{\beta}{12}\zeta^2+\cdots}$$ Now this can in principle be inverted. Linear term is obvious. The rest, you can get by Taylor expanding the right hand side and then applying something like the Lagrange inversion formula,or, better still, just by recursively plugging the result beck into itself.
Taylor expansion of the right hand side by the conventional binomial theorem:
$$i\delta=\zeta\left[1+\frac12\left(\frac{\alpha}{3}\zeta+\frac{\beta}{12}\zeta^2+\cdots\right)-\frac{1}{8}\left(\frac{\alpha}{3}\zeta+\frac{\beta}{24}\zeta^2+\cdots\right)^2+\cdots\right]=$$ $$=\zeta+\frac{\alpha}{6}\zeta^2+\left(\frac{\beta}{24}-\frac{\alpha^2}{72}\right)\zeta^3+\cdots$$
Express $\zeta$ and you have a function: $$f(\zeta)=i\delta-\frac{\alpha}{6}\zeta^2-\left(\frac{\beta}{24}-\frac{\alpha^2}{72}\right)\zeta^3+\mathcal{O}(\zeta^4)$$ You are looking for $\zeta=f(\zeta)$, which you get by recursion (don't worry, the Taylor series you get is stable, you just can't trust the terms higher than what you have in the original series, in that case, third order).
First iteration:
$$f(0)=i\delta$$ second (still wrong at the red part): $$f(i\delta)=i\delta+\frac{\alpha}{6}\delta^2+\left(\frac{\beta}{24}-\color{red}{\frac{\alpha^2}{72}}\right)i\delta^3$$ third: $$f(f(i\delta))=i\delta+\frac{\alpha}{6}\delta^2+\left(\frac{\beta}{24}-\frac{5\alpha^2}{72}\right)i\delta^3+\mathcal{O}(\delta^4)$$
You can also just use the method of undetermined coefficients.