In Griffith's Quantum Mechanics book, when he tackles time-independent perturbation theory, he states that one can write the eigenfunctions of the perturbed system, $$ H = H^0 + \lambda H', $$ as a power series in $\lambda$, $$ \psi_n = \psi_n^0 + \lambda\psi_n^1 + \lambda^2\psi_n^2 + \cdots $$ I'm trying to understand this intuitively. I am guessing that this is some way of writing the Taylor expansion of $\psi_n$. Then I do indeed understand that summing up all the terms gives us the full eigenfunction to arbitrary precision.
I just can't see how the power series in the book is really a Taylor expansion. Given that this is time-independent, the eigenfunctions should be functions of $x$, and their general Taylor expansion should look like $$ \psi_n(x) = \psi_n(a) + \psi_n'(a)(x - a) + \frac{\psi_n''(a)}{2!}(x - a)^2 + \cdots. $$ He writes that $\psi_n^1$ is the first-order correction to the $nth$ eigenfunction, $\psi_n^2$ the second-order, and so on. Can this statement be reconciled with writing the corrections as $$ \psi_n^m(a) = \frac{1}{m!}\frac{d^m\psi_n(x)}{dx^m}\bigg\rvert_{x=a}, $$ such that the Taylor expansion becomes $$ \psi_n(x) = \psi_n^0(a) + \psi_n^1(a)(x - a) + \psi_n^2(a)(x - a)^2 + \cdots, $$ or what? I would still need to define something like $$ x - a \equiv \lambda $$ to get $$ \psi_n(x) = \psi_n^0(a) + \lambda\psi_n^1(a) + \lambda^2\psi_n^2(a) + \cdots, $$ but this forces some dependencies on $\lambda$ that isn't physically motivated, and screws up the relations between $x$, $a$ and $\lambda$ in the equation, so it seems like a stupid thing to do.
