Analysis of the eigenvalues of the particle in a finite square well The eigenstates of the particle in a 1D finite square well Hamiltonian:
\begin{align}
    H = \frac{\hat{p}^2}{2m} + V(x)
\end{align}
\begin{align}
    V(x) = 
    \begin{cases}
        -V_0 & \text{if } |x| < a \\
        0 & \text{otherwise}
    \end{cases}
\end{align}
belong to eigenvalues given by the following expression:
\begin{align}
    E = \frac{\hbar^2 \kappa^2}{2 m a^2} - V_0
\end{align}
Where $\kappa$ is any allowed solution of the following equation:
\begin{align}
    \tan{\kappa} = \sqrt{\left(\frac{\kappa_0}{\kappa}\right)^2 - 1}
\end{align}
and $\kappa_0$ is given in terms of the parameters of the problem. One can show by plotting $\tan{\kappa}$ and $\sqrt{\left(\frac{\kappa_0}{\kappa}\right)^2 - 1}$ on the same axes and reasoning about the limit of the intersection points between the two curves as $V_0$ tends to infinity that the limit of the eigenvalues are (half of) those for the infinite square well.
It would be useful to obtain a more general approximate expression for the eigenvalues and it seems that such an expression could be obtained by

*

*Inserting a small parameter into the equation for $\kappa$


*Expanding the allowed values of $\kappa$ as a perturbation series in a small parameter $\epsilon$ and about the limiting solutions


*Substituting the result into the equation for $\kappa$ and expanding both sides carefully as taylor series in $\kappa$


*Solving the resulting equation for the perturbation series coefficients by requiring both sides of the equation to be equal term-by-term in powers of $\epsilon$ (and if I'm lucky, summing the series in $\epsilon$
However, this seems undesirable for two reasons

*

*It would be nice to be able to reason about the limiting values of $\kappa$ in a more concrete way. Ideally, one could use some kind of asymptotic representation of the solutions to do so, but this cannot be possible if the perturbation series is taken about the asymptotic solutions to start with


*It is unlikely to be possible to sum the series in $\epsilon$ and therefore, limited intuition can be gained about the form of the solution. It is also for this reason that numerical solutions are undesirable
Therefore, my question is this: is there some other technique for obtaining an approximate expression for ALL the allowed values for $\kappa$? Is it the case that we can't really do better than a numerical solution?
 A: I’m not sure what you’re trying to do here.  The eigenvalues are solutions to the transcendental equation
$$
\tan(\kappa)=\sqrt{(\kappa/\kappa_0)-1}
$$
so the best way to solve this is to get a guess by looking at approximate intersections of the curves, followed by a simple Raphson-Newton finding of roots using the guesses as starting points.  It’s basically a numerical approach.
I suppose in some sense Raphson-Newton is equivalent to a series expansion, but
the kind of series summation you suggest is unlikely to work else it would be all over the textbooks by now: in any case, as there are multiple eigenvalues satisfying the same transcendental equations, the resumming is bound to depend quite delicately on the initial value.
You might care to consult

Flügge, S., 2012. Practical quantum mechanics. Springer Science & Business Media.

where some sort of approximate root finding is discussed (at least for the lowest few eigenvalues).  I don’t remember if Flugge gets his transcendental in the same form as yours, but it’s definitely in there
