If we add quartic term in quantum harmonic oscillator,
$$V(x)=\frac{mx^2}{2}+\frac{m^{2}\omega^{3}}{\hbar}\hat{x}^{4}.$$
$$H(\lambda)\,=\,H^{(0)}+\lambda\,\frac{m^{2}\omega^{3}}{\hbar}\dot{x}^{4}\,=\,H^{(0)}+\lambda\,\frac{1}{4}\,\hbar\omega(\dot{a}+\hat{a}^{\dagger})^{4}.$$
Using time independent perturbation theory, $$E_{0}(\lambda)=\frac{_1}{^2}\hbar\omega\left(1+\frac{^2}{2}\lambda-\frac{21}{4}\lambda^{2}+\frac{33}{8}\lambda^{3}-\frac{3585}{64}\lambda^{4}+\frac{916731}{128}\lambda^{5}-\frac{65518401}{812}\lambda^{6}+O(\lambda^{7})\right).$$
In MIT quantum Physics III course notes it is written that
As it turns out the coefficients keep growing and the series does not converge for any nonzero $\lambda$; the radius of convergence is actually zero! This does not mean the series is not useful. It is an asymptotic expansion. This means that for a given small value of $\lambda$ the magnitude of successive terms generally decrease until, at some point, they start growing again. A good approximation to the desired answer is obtained by including only the part of the sum where the terms are decreasing.
I am not able to understand that how perturbative expansion is useful there. It says that for any value of $\lambda$, after some point the successive terms starts increasing. Then how the perturbative solution restricted to some order is useful. The main aim of perturbative expansion is that if $\lambda$ is small then the successive terms in the series starts decreasing and we can neglect them and the solution up to first or second order is a good approximate solution.
But how restricting the number of terms to the point where the terms starts increasing gives us the good approximate solution here as written in the notes? How can we neglect the higher order terms then? I think I am missing something.
Please explain.