first inequality is similar. Notice that for all $n\in[1,y/H]$ we have $A_{n,y}\subset\{(r_1,\dots,r_y)\in{\mathbb{N}}^y\,:\,
\sum_{i=1}^{[y/H]}r_i\ge H(y/H)\}$. Again by Corollary \[corexp\] for ${\widetilde}{\Pi}_y^{\varrho}$ and $y/H$ in place of $\theta n$ we get $$e^{-c(H)y/H}\ge \sum_{A_{n,y}} \prod_{i=1}^y
{\varrho}(1-q)^{r_i}F_M(r_{i-1}, r_i, u_{i-1}),\quad  \text{for all $n\in[1,y/H]$}.$$ The rest of the proof follows the proof of the second inequality.

\[lemfinal\] Let $h$ be as in Lemma \[corbond\]. There exists a strictly positive $c_2=c_2({\varrho},M)$ such that for $y$ large $$(1-q)^y\sum_{1\le n< y/h}\sum_{A_n}
  \prod_{i=1}^n {\varrho}F_M(r_{i-1}, r_i, u_{i-1}) \geq c_2.$$

We start with (\[nn\]) and sum up over $n\in[y/h,y]$. The number of terms in this summation does not exceed $y$, therefore, for all sufficiently large $y$ the sum is less than $(1-e^{-M})/2$. Since the sum over $n\in[0,y)$ is equal to $\Pi^{\varrho}_y({\mathbb{N}}^y)\ge
  (1-e^{-M})$, we conclude that $$\frac12\,(1-e^{-M})\le
  \sum_{1\le n< y/h}\sum_{A_{n,y}} \prod_{i=1}^y {\varrho}(1-q)^{r_i}F_M(r_{i-1},
  r_i, u_{i-1}).$$ Just as in the previous proof, if $y\ge n+1$ then we perform first the summation over $r_{n+1},\dots,r_y\in{\mathbb{N}}$ and apply (\[co2\]) to get $$\frac12\,(1-e^{-M})\le
  (1-e^{-M})^{-1}\sum_{1\le n< y/h} \sum_{A_{n,n}} \prod_{i=1}^n
  {\varrho}(1-q)^{r_i}F_M(r_{i-1}, r_i, u_{i-1}).$$ Next, we replace $F_M(r_{n-1},r_n,u_{n-1})$ with its upper bound $F_M(r_{n-1},y-\sum_{i=1}^{n-1}
  r_i,u_{n-1})$ and sum over $r_n$ from $y-\sum_{i=1}^{n-1} r_i$ to infinity. We obtain $$\begin{gathered}
    \frac12\,(1-e^{-M})\le \frac{1}{q(1-e^{-M})}\,\sum_{1\le n< y/h}
    \sum_{A_n} \prod_{i=1}^n {\varrho}(1-q)^{r_i}F_M(r_{i-1}, r_i,
    u_{i-1})\\=\frac{(1-q)^y}{q(1-e^{-M})}\,\sum_{1\le n< y/h}
    \sum_{A_n} \prod_{i=1}^n {\varrho}F_M(r_{i-1}, r_i, u_{i-1}).
  