that $\lim_{l\rightarrow\infty} (\overline{r}(p_{n_l})-\overline{r}(z_{n_l}))=0$. Indeed: let $$\overline{r}(z)=mz+b_1z^{\frac{s-s_1}{s}}+b_2z^{\frac{s-s_2}{s}}+\cdots,\,\,\, s, s_i\in {\Bbb N},\, 0<s_1<s_2<\cdots.$$ Thus, $$\overline{r}\,'(z)={m} +\frac{s-s_1}{s}b_1z^{\frac{-s_1}{s}}+\frac{s-s_2}{s}b_2z^{\frac{-s_2}{s}}+\cdots\,_{\overrightarrow{z\rightarrow
\infty}} \,{m}.$$ Therefore, there exist $K>0$ and $\delta>0$ such that $|\overline{r}\,'(z)|\leq K$, for $|z|>\delta$. Applying the Mean Value Theorem (see [@Ahlfors]), we have that $$\Re\left(\frac{\overline{r}(p_{n_l})-\overline{r}(z_{n_l})}{p_{n_l}-z_{n_l}}\right)=\Re (\overline{r}\,'(c_1)),\qquad
\Im\left(\frac{\overline{r}(p_{n_l})-\overline{r}(z_{n_l})}{p_{n_l}-z_{n_l}}\right)=\Im (\overline{r}\,'(c_2)),$$ where $\Re(q)$ and $\Im(q)$ denote the real part and the imaginary part of $q(z)\in {\Bbb C}(z)$, respectively, and $c_1, c_2\in ]p_{n_l}, z_{n_l}[$, where $]p_{n_l}, z_{n_l}[:= \{z \in {\Bbb C}:\, z =p_{n_l}+ (p_{n_l} - z_{n_l})t,\, t\in (0,1)\}.$ Thus, $$|\overline{r}(p_{n_l})-\overline{r}(z_{n_l})|^2=(\Re (\overline{r}\,'(c_1))^2+\Im (\overline{r}\,'(c_2))^2)|p_{n_l}-z_{n_l}|^2.$$ Now, since $\lim_{l\rightarrow\infty}p_{n_l}=\lim_{l\rightarrow\infty}z_{n_l}=\infty$, and $\lim_{l\rightarrow\infty}p_{n_l}-z_{n_l}=0$, we deduce that given $\varepsilon>0$, there exists $l_1\in {\Bbb N}$ such that, for $l\geq l_1$, $$|p_{n_l}|>\delta+\varepsilon,\quad |z_{n_l}|>\delta+\varepsilon, \quad \mbox{and}\quad
|p_{n_l}-z_{n_l}|<\varepsilon.$$ Then, $|c_j|>\delta$ and $|\overline{r}\,'(c_j)|\leq K$ for $j=1,2$, which implies that, for $l\geq l_1$, $$|\overline{r}(p_{n_l})-\overline{r}(z_{n_l})|\leq \sqrt{2}K|p_{n_l}-z_{n_l}|_{\overrightarrow{l\rightarrow
\infty}}0$$ (note that $\Re (\overline{r}\,'(c_1))\leq |\overline{r}\,'(c_1)|\leq K$, and $\Im (\overline{r}\,'(c_2))\leq |\overline{r}\,'(c_2)|\leq K$). Therefore, $\lim_{l\rightarrow\infty} (\overline{r}(p_{n_l})-\overline{r}(z_{n_l}))=0$, which implies that there exists a sequence $\{z_{n_l}\}_{l\in {\Bbb N}}$ with $\lim_{l\rightarrow\infty} z_{n_l}=\infty$, such that $$\lim_{l\rightarrow\infty}(r(z_{n_l})-\overline{r}(z_{n_l}))=0$$ (see inequality (I)). Then, the terms with positive exponent of the series $r(z)$ and $\overline{r}(z)$ are the same (see the proof of Lemma \[L-DistVertical\]). Hence, we conclude that (see Lemma \[L-DistVertical\]) $$\lim_{z\rightarrow\infty}(r(z)-\overline{r}(z))=0$$ and thus $B$ and $\overline{B}$ are convergent (see Definition \[D-distance0\]).

Reciprocally, let us assume that 