[@Be-Vo].

The Boolean case is easier. (\[uplus\]) implies that $$F_{\mu_N ^B} (z) = F_{\mu}(Nz) - (N-1)z \to F_{\mu_{a,b}}(z) = z - m_1(\mu) ~~\text{as}~ N \to \infty.$$ The weak convergence follows from Theorem 2.5 in [@Maa].

We can prove a similar result for the monotone convolution. This is quite new since there are not complex analytic methods to analyze iteration of monotone convolutions of probability measures without the usual moments. Now we prove a limit theorem using the complex moments. In this case we restrict the class of probability measures to $\mathcal{P}_1$. We need a Lemma.

\[lem1\] If $|m_n(\mu)| \leq R^n$ for all $n \geq 0$, then $|m_n(\mu^{\rhd N})| \leq (NR)^n$ for all $n,~N \geq 0$.

We prove this by induction. Assume that $|m_n(\mu^{\rhd N})| \leq (NR)^n$ for all $n \geq 0$. Since we complex moments via analytic continuation, it holds that $$\label{moment}
\begin{split}
m_n(\rho \rhd \sigma) &= \sum_{k = 0} ^{n} \sum_{\substack{j_0 + j_1 + \cdots +j_k = n - k, \\  0 \leq j_l, ~0 \leq l \leq k }} m_k(\rho) m_{j_0}(\sigma)\cdots m_{j_k}(\sigma)   
\end{split}$$ for $\rho,~\sigma \in \mathcal{P}_1$ [@H-S]. If $|m_n(\rho)| \leq R_1^n$ and $|m_n(\sigma)| \leq R_2^n$ for all $n$, then we have $$\notag
\begin{split}
|m_n(\rho \rhd \sigma)| &\leq 