subadditive. From this, (\[us\]), and (\[un\]) it follows that $$\log
  \mu= \lim_{n\to\infty} \dfrac{\log {\widetilde}{U_n}}{n}=\inf_n\frac{\log
    {\widetilde}{U_n}}{n}.$$ We conclude that ${\widetilde}{U_n}\ge \mu^n$. By (\[us\]), $U_n \ge \mu^n(1-e^{- M}) $ for all $n\in{\mathbb{N}}$.

Properties of $\mu=\mu(q)$ follow from the inequality $U^{1/n}_n\le \mu\le
  U^{1/n}_n(1-e^{-M})^{-1/n}$, $n\in{\mathbb{N}}$. Taking $n=1$ we compute directly that $$U_1(q)=\sum_{r=1}^\infty
  q(1-q)^{r-1}/(2e^Mr)=\frac{q\log(1/q)}{2e^M(1-q)}$$ and obtain the desired bounds on $\mu$. Continuity of $\mu(q)$ follows from continuity of $U^{1/n}_n(q)$ as a function of $q$ for each $n$ and the fact that $(1-e^{-M})^{-1/n}\to 1$ as $n\to\infty$.

\[corbase\] For each ${\varrho}\in(0,\infty)$ there is a $q=q({\varrho},M)\in(0,1)$ such that $$\label{rho}
    2(1-e^{-M})\le
  e^{-M}{\varrho}\log (1/q)\le 2,$$ and for all $n\in{\mathbb{N}}$ $$\label{co1}
  (1-e^{-M})\le \sum_{r_1,r_2,\dots,r_n\in{\mathbb{N}}} \prod_{i=1}^n
\left((1-q)^{r_i}{\varrho}F_M(r_{i-1}, r_i, u_{i-1})\right)\le 1.$$ Moreover, for every $m\in\{1,2,\dots,n\}$ and any $r_1,r_2,\dots,r_{m-1}\in{\mathbb{N}}$ $$\label{co2}
  (1-e^{-M})\le \sum_{r_m,r_{m+1},\dots,r_n\in{\mathbb{N}}} \prod_{i=m}^n
\left((1-q)^{r_i}{\varrho}F_M(r_{i-1}, r_i, u_{i-1})\right)\le (1-e^{-M})^{-1}.$$

Lemma \[lem5basic\] implies that $$(1-e^{- M})\le \sum_{r_1,r_2,\dots,r_n\in{\mathbb{N}}} \prod_{i=1}^n
  (1-q)^{r_i}\dfrac{q}{(1-q)\mu} F_M(r_{i-1}, r_i, u_{i-1}) \le 1.$$ Setting ${\varrho}= q/(\mu(1-q))$ we get (\[co1\]). Properties of $\mu$ (see (\[mu\])) imply (\[rho\]).

To show (\[co2\]) we first notice that applying (\[co1\]) to the walk which starts at $x_{m-1}$, never returns to $x_{m-1}$, and reaches $x_n$ (in the notation of Lemma \[F\]) we get $$\begin{gathered}
\label{red}
    (1-e^{-M})\le \sum_{r_m,r_{m+1},\dots,r_n\in{\mathbb{N}}}
    \Big((1-q)^{r_m}{\varrho}F_M(0, r_m, 0)    \prod_{i=m+1}^n (1-q)^{r_i}{\varrho}F_M(r_{i-1}, r_i,
        u_{i-1})\Big)\le 1.
  \end{gathered}$$ Moreover, by (\[defF\]), $$F_M(0, r_m, 0)\le F_M(r_{m-1}, 