This is again a question in the context of this paper about the Exact Renormalization Group. On p 23 and the following few pages, it is explained that for a $\lambda \phi^4$ bare action at the bare scale $\Lambda_0$, after integrating out degrees of freedom and assuming a small coupling $\lambda$, the effective action at the larger scale $\Lambda$ can be written as the sum of a perturbative series plus nonperturbative power corrections (Eq. 2.6)
$$ S_{\Lambda,\Lambda_0}[\phi] = \sum\limits_{i=0}^{\infty}\lambda^{i-1}S_i[\phi] +O(\Lambda/\Lambda_0) $$
Naively taking $\Lambda_0 \rightarrow \infty$ makes the second part disappear and leaves the first part which is self-similar which means that the theory is renormalizable. However, as stated in the paper, UV renormalons can make the perturbation series ambiguous, such that the power corrections and therefore the bare scale can not be removed and the theory would then not be renormalizable.
The issue is explained by the mathematical argument that in order to obtain a unique finite value for the perturbative series, one makes use of the so-called Borel transformation to define a function which has the same power series expansion defined as an integral in the complex Borel plane. This integral exists only of there are no poles on the real axis, otherwise the contour of integration around these poles can be deformed which makes the integral and the function defined by it ambiguous. If this is the case, the power correction terms have to be kept to restore the uniqueness of the integral which means that the bare scale can not be removed.
In the paper it is said, that the poles on the real axis are for example due to UV renormalons, that arise from large loop momenta in certain Feynman diagrams.
My question now is:
What are these renormalons from a physics point of view? How do they enter the Lagrangian of the theory? Are they some kind of unphysical auxilliary fields that appear in the mentioned Feynman diagrams? And what do the Feynman diagrams that contain them look like?