In his paper [Constructive Renormalization Theory][1], V. Rivasseau describes the idea of Wilson's approach of solving path integrals step by step. In section 1.4, page 5, however, there is a statement which I do not follow. He discusses a $\phi^{4}$ theory and decomposes the covariance into a sum:
\begin{eqnarray}
C_{0}(p) = \int_{1}^{\infty}e^{-\alpha(p^{2}+m^{2})}d\alpha \quad \mbox{and} \quad C^{j}(p) = \int_{M^{-2j}}^{M^{-2(j-1)}}e^{-\alpha(p^{2}+m^{2})}d\alpha \tag{1}\label{1}
\end{eqnarray}
in such a way that:
\begin{eqnarray}
C_{\rho}(p) = \sum_{j=0}^{\rho}C^{j}(p) \tag{2}\label{2}
\end{eqnarray}
He proceeds to explain how the partition function can be obtained by performing interativelly $\rho+1$ (convolution) integral. At some point, he states the following:

> We see that constructing the ultraviolet limit is the same as finding a bare action $S_{\rho}(\phi)$ such that the effective action, or renormalized action $S_{0}(\phi)$ converges as $\rho \to \infty$. 

I might be missing something, but I don't fully understand this statement. First, it seems that, in the limit $\rho \to \infty$ one recovers the the *regularized* theory, because (\ref{1}) is a telescoping series with UV cutoff and as $j \to +\infty$ one ends up with regularized covariance
\begin{eqnarray}
C_{0}(p) = \frac{1}{p^{2}+m^{2}}e^{-(p^{2}+m^{2})}. \tag{3}\label{3}
\end{eqnarray}
In other words, I don't see how it is possible to take the UV limit using this decomposition into steps as a limiting case. Second, it is a little bit odd for me to say that one finds $S_{\rho}(\phi)$ so that $S_{0}(\rho)$ exists when $\rho \to \infty$. The flow is going backwards, from $\rho$ to zero, how can you then $\rho \to \infty$ afterwards? Is it just because $S_{0}$ should depend on $\rho$ somehow?

  [1]: https://arxiv.org/pdf/math-ph/9902023.pdf