The ambiguity that people normally refer to is due to the lack of Borel summability of the perturbation series.
Consider a series of the form
$$A(g) = \sum_{n=1}^{\infty}{(-1)^n g^n (n!)} $$
If the coefficients $b_n$ are of order $1$, this series is obviously divergent. But we can compute its Borel sum. First compute the Borel transform:
$$\mathcal{B}A(t) = \sum_{n=1}^{\infty}{ (-1)^n t^n }$$
We may then compute the Borel sum:
$$ S(g) = \int_0^{\infty}{dt\, e^{-t} \mathcal{B}A(t g) }= \int_0^{\infty}{dt\, e^{-t} \sum_{n=1}^{\infty}{(-1)^n (tg)^n }}$$
We now pull the sum out of the integral, assuming the series can be integrated term by term. This, technically speaking, is a wrong thing to do since this series is not uniformly convergent -- in fact, it's not convergent at all -- but we'll do it anyway with the proviso that we'll simply define the answer to be what we get by doing this. I'm the one defining how to sum up this series, so I'm entitled.
$$S(g) = \int_0^{\infty}{dt\, \frac{e^{-t}}{1 + tg} } $$
And as long as $g > 0$, this expression is obviously finite. The procedure worked and we were able to find a nice well defined number to assign to this series.
You'll notice however that the fact that the original series was alternating was crucial for obtaining this result. Let's repeat it for a non alternating series, which is more representative of the kinds of series you might find when adding up Feynman diagrams in a field theory.
$$S(g) = \int_0^{\infty}{dt\, e^{-t} \sum_{n=1}^{\infty}{ (tg)^n }} = \int_0^{\infty}{dt\, \frac{e^{-t}}{1 - tg} } = \frac{1}{g} \int_0^{\infty}{dt\, \frac{e^{-\frac{t}{g}}}{1 - t} }$$
That is, for $g > 0$ the integrant has a pole on the positive real axis. You can still do the integral but you must choose which in which direction you must do your excursion into the complex plane. This is an ambiguity.
Notice that the residue at the $t = 1$ pole in the above is
$$Res(S,1) = -\frac{e^{-\frac{t}{g}}}{g} $$
Curiouser and curiouser! The imaginary part you find in the Borel sum is nowhere to be found in the original series -- since every term is real -- and, further, it is non perturbative in $g$! Hmm... that's suggestive.
It turns out that if you perform your calculations carefully enough and any analytic continuations are carried out consistently, then the imaginary contributions associated with instantons cancel out the imaginary contributions from the failure of the perturbation series to be Borel summable. The full expression for the energy eigenvalues therefore suffers from no ambiguities. This phenomenon where high orders in perturbation theory (about the trivial vacuum) somehow encode information about low orders around instanton solutions has been termed "resurgence". Key words in the literature also include "trans series", "Stokes wedges" and "Lefschetz thimbles".
Refs:
http://arxiv.org/abs/1210.2423
http://arxiv.org/abs/1210.3646
http://arxiv.org/abs/1411.3585
