Is there a counterexample to this pattern in QM perturbation theory? Consider some $1d$ system $H_0$ with energy levels $E_n^{(0)}$ that has been perturbed by $\lambda V$, i.e. $H = H_0 + \lambda V$ for some $\lambda >0$. Consider the ratio of the first order perturbation $E_n^{(1)}$ to the unperturbed $E_n^{(0)}$ as a function of $n$.
If this ratio $E_n^{(1)}/E_n^{(0)}$ diverges as $n\to \infty$, must the perturbation series fail to converge in $\lambda$?

Here's what a counterexample might look like: For example, suppose the true energy levels of the system were $n^q e^{-\lambda n^p}$ for $q, p>0$. Then the first order perturbation would be $-\lambda n^{p+q}$, which would suffer $E_n^{(1)}/E_n^{(0)} \to \infty$ as $n \to \infty$. Nevertheless, the form of the energy levels is analytic in $\lambda$ and hence one would expect the perturbation series to converge.
My trouble is that I'm unsure of how to construct a counterexample Hamiltonian with such energy levels. Indeed, many examples where $\lim_{n\to \infty} E_n^{(1)}/E_n^{(0)} = \infty$ in fact do have perturbation series that fail to converge for any nontrivial $\lambda$! For example, the Harmonic oscillator perturbed by $\lambda x^k$ for even integers $k>2$  suffers a first order correction growing strongly in $n$ and for which $\lim_{n\to \infty} E_n^{(1)}/E_n^{(0)} = \infty$, and these perturbed systems are known to have energies that are not analytic in $\lambda$ about $\lambda = 0$ by Dyson's argument (and in turn have merely asymptotic and not convergent perturbation series).

I'm curious if such a counterexample is known/can be constructed or if there's some friendly argument why $\lim_{n\to \infty} E_n^{(1)}/E_n^{(0)} = \infty$ completely ensures the perturbation series cannot converge in $\lambda$. To avoid any subtleties with negative energy bound states approaching an energy of $0$, let's restrict to the case of $E_n^{(0)} \to \infty$ as $n \to \infty$.
 A: The limit $n\rightarrow \infty$ is unnecessary (as is looking for a divergence in this ratio). A condition that is typically necessary for perturbation theory to be valid is that
\begin{equation}
\frac{E^{(1)}_n}{E^{(0)}_n} \ll 1
\end{equation}
for all $n$. (At least, all $n$ where you trust your theory is valid).
An exception is if $E_n^{(0)}=0$, then the leading contribution to the energy can the first order correction. But then you would expect the second order correction to be smaller than the first.
A: The first order correction is the diagnonal element of pertirbation:
$$
E_n^{(1)}=\langle n|\lambda V|n\rangle,
$$
which is why it is usually absorbed in the unperturbed Hamiltonian
$$
E_n^{(0)}=\langle n|H_0|n\rangle \longrightarrow \tilde{E}_n^{(0)}=\langle n|H_0|n\rangle + \langle n|\lambda V|n\rangle
$$
The rest of the perturbation terms contains denominators inversely proportional to energy differences
$$
\sim\frac{1}{E_n^{(0)}-E_m^{(0)}}.
$$
The condition $$E_n^{(1)}/E_n^{(0)}$$ might make the diagonal part of the Hamiltonian somehow irregular - e.g., by introducing degeneracies that would break perturbation theory, $\tilde{E}_n^{(0)}\approx \tilde{E}_m^{(0)}$ - it depends on the details of the limit $n\rightarrow 0$, and no general answer can be given. However, as should be clear from the above, this is more an issue of a poorly defined unperturbed Hamiltonian, than an actual pattern in perturbation theory.
Example
Let us perturb a harmonic oscillator with unperturbed energies
$$
E_n = \hbar\omega_0 n
$$
We now take a perturbation with diagonal elements
$$
\langle n|V|n\rangle = V_0n^2
$$
If $V_0>0$ the distances between the levels would only grow with increasing $n$, and the perturbation theory works fine, as long as the non-diagonal elements of the perturbation is smaller than the distance between the levels. However, if $V_0<0$, we might get level crossings, i.e., situations where the levels are nearly degenerate and the PT breaks down.
