How to carry out the perturbation expansion of an anharmonic oscillator to high orders? I think this is a standard problem in quantum mechanics. Consider the anharmonic oscillator 
$E \psi = \left(- \frac{1}{2} \frac{\partial^2}{\partial^2 x } + \frac{1}{2}x^2 + \epsilon x^4 \right) \psi$.
Formally, the ground state energy has the asymptotic expansion 
$E(\epsilon) \sim  \sum_{n=0}^\infty a_n \epsilon^n$.
How to calculate the coefficients $a_n$ to high orders, say, for $n= 20$?
 A: As mentioned in the comments by Bubble, this is answered in

Ground State Energy Calculations for the Quartic Anharmonic Oscillator, Robert Smith. Notes for Math 4901, University of Minnesota, Morris (2013).

but as the document is not crawlable by the Wayback Machine I'll summarize it here.
Smith considers hamiltonians of the form
$$
H=-\frac12 \frac{\mathrm d^2}{\mathrm dx^2}+\frac12 x^2+\lambda x^{2K}
$$ 
for $K=2,3,4,\ldots$, and the resulting spectrum is of the form
$$
E(\lambda)=\sum_{i=0}^\infty E_i \lambda^i,
$$
where the coefficients are given as
$$
E_i=\frac1iX_{K,i-1},
$$
for $i\geq1$, in terms of the variables $X_{K,i}$ which satisfy the recurrence relations
$$
X_{j,i}=\frac1{2j}\left\{
\frac{j-1}2 \left[4(j-1)^2-1\right]X_{j-2,i}+2(2j-1)\sum_{m=0}^iE_mX_{j-1,i-m}-2(2j+k-1)X_{j+k-1,i-1}
\right\}
$$
and
$$
X_{j,0}=\frac{2j-1}j E_0 X_{j-1,0}+\frac{j-1}{4j}(4j^2-8j+3)X_{j-2,0}
$$
with initial conditions $X_{0,i}=\delta_{0,i}$.
The first few terms of this expansion are
$$
E(\lambda)=\frac12+\frac34\lambda-\frac{21}8\lambda^2+\frac{333}{16}\lambda^3-\frac{30885}{128}\lambda^4+\cdots.
$$
This result is obtained via the method of Swenson and Danforth, which Smith explains in detail in Appendix A, with a further reference to the method in

Fernandez, Francisco. Introduction to Perturbation Theory in Quantum Mechanics. New York, CRC Press, 2001. 


(Personally, though, I tend to think that if you're getting this intense about an $x^4$ anharmonicity, then you should also be worrying about terms in $x^6$ - and that's in an ideal case with no symmetry-breaking odd terms. But that's just me.)
