sum of such diagrams is also finite (in the large $N$ limit) in the non-supersymmetric theory at a given order in perturbation theory. To see this consider the correlator $\left< ({\ensuremath{\mathcal{O}}}_1)^3 \right>$ in the ${\ensuremath{\mathcal{N}}}=2$ theory, which does not contain divergences since both ${\ensuremath{\mathcal{O}}}_1$ and the $(\phi^\dagger\phi)^3$ coupling are not renormalized in the ${\ensuremath{\mathcal{N}}}=2$ theory. The diagrams contributing to this correlator again factorize into a product of matter loops, that are in general of a lower order in perturbation theory (the only difference is that the diagrams may now include both $\bar{\psi}\psi\bar{\phi}\phi$ and $(\phi^\dagger\phi)^3$ vertices). The proof then follows in a similar way.

We are now ready to show that $\beta_{\lambda_6} = O(1/N)$. In our non-supersymmetric model, at large $N$ the correlator $\left< ({\ensuremath{\mathcal{O}}}_1)^3 \right>$ receives contributions from two types of diagrams, with either zero or one $(\phi^\dagger\phi)^3$ vertices:

  ---------------------------------- ----------------------------------
   ![image](G1.pdf){height="4.5cm"}   ![image](G2.pdf){height="4.5cm"}
                 (G1)                               (G2)
  ---------------------------------- ----------------------------------

The $\lambda_6$ coupling contributes to $\left< ({\ensuremath{\mathcal{O}}}_1)^3 \right>$ at large $N$ through diagrams of the form (G2), some of which are non-zero (such as the leading order diagram which is explicitly drawn). Thus, if the beta function is non-zero we must have divergences in $\left< ({\ensuremath{\mathcal{O}}}_1)^3 \right>$. However, we have shown above that 