What does it mean for a perturbation expansion to break down? For an anharmonic oscillator with Hamiltonian $H = {\hat p_x^2\over 2m}+{1\over 2}m\omega^2\hat x^2+b\hat x^4$ I found the first order shift in energy is:
$$E_n^{(1)}={3 \hbar^2 b\over 4m^2\omega^2}\big(1+2n+n^2\big)$$
Now I need to argue that no matter how small b is, the perturbation expansion will break down for sufficiently large $n$. I know that the energy will keep increasing as $n$ increases because $b$ is a constant, but I don't know what it means for a "perturbation expansion to break down." What does this mean?
 A: The basic assumption behind perturbation theory is that $E_n^{(1)}$ should be “much smaller” than the unperturbed energy $E_n^{(0)}$.  This assumption clearly is no longer value for sufficiently large $n$ irrespective of $b$ since the correction grows like $n^2$ while the unperturbed energy grows like $n$.  There is no “hard and fast” rule as to when perturbation theory break down because “much smaller” is rather qualitative, but you might look at the point where corrections are - say - $20\%$ or more of the unperturbed energy as a point where the correction is no longer small.  
You might also look at the first correction to eigenstate: if the unperturbed state is not “significantly more” probable than the other states in the perturbed eigenstate then perturbation theory can be said to be invalid.
A: In a perturbative series, you want successive corrections to be smaller than the previous ones, so that you get a convergent result. This breaks down if higher-order shifts get larger, which you can readily see will happen as $n$ grows large. 
A: Suppose that you want to compute corrections to some observable $f_0$:
$$f(x) = \sum_{k \leq N} f_k \, x^{k} + O(x^{N+1}).$$
This is a good estimate only if the tail
$$\lVert \, f(x) - \sum_{k \leq N} f_k \, x^{k} \rVert = \lVert \, f_{N+1} x^{N+1} + \ldots \rVert$$
is small. If the first error term $f_{N+1} x^{N+1}$ is of the same order as the LHS, the tail is probably large and you're not doing something sensible. In your case $N=0$, but you can still get a feeling if perturbation theory will work by requiring that $E_n^{(1)} \ll E_n^{(0)}$.
