Error in perturbation series question

[AxiomOfChoice]

Suppose we know the perturbation series

<br /> E = E(\epsilon) = E_0 + \epsilon E_1 + \epsilon^2 E_2 + \ldots<br />

converges, where E_0 is a discrete eigenvalue of H_0 and we are considering a Hamiltonian H = H_0 + \epsilon H_1. Does this mean that we know

<br /> E - E_0 = O(\epsilon)<br />

as \epsilon \to 0 in the precise sense that we know there exists a \delta &gt; 0 and a C &gt; 0 such that if |\epsilon| &lt; \delta, then |E - E_0| \leq C|\epsilon|?

 

[mfb]

If the perturbation series converges, this C has to exist - that is just mathematics (taylor series).

 

[The_Duck]

If you really have a convergent series, then I think your statement holds just as a general property of power series, and you can pick any $C > |E_1|$. As I understand it, however, perturbation series typically do *not* converge. They are usually asymptotic series, meaning that if you truncate them after order $\epsilon^N$, the error between the truncated series and the actual eigenvalue vanishes like $\epsilon^{N+1}$ as $\epsilon \to 0$ (which statement you can formalize if you like in the same epsilon-delta style).