$\newcommand{\al}{\alpha} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\om}{\omega} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}} \newcommand{\tf}{\widetilde{f}}$

The answer depends on whether the support (say $S$) of the distribution of $\xi_1$ contains $1$ or not. If $1\in S$, then the answer is no; otherwise, yes.

Indeed, suppose first that $1\in S$. For each natural $N$, let \begin{equation} A_N:=\Big\{\forall k \ge 0\ \prod_{i=1}^{N}\xi_{i+kN} > 6N\Big\}=\bigcap_{k=0}^\infty B_{N,k}, \end{equation} where \begin{equation} B_{N,k}=\Big\{\prod_{i=1}^{N}\xi_{i+kN} > 6N\Big\}. \end{equation} The events $B_{N,0},B_{N,1},\dots$ are independent and for each $k$ \begin{align*} \P(B_{N,k})=\P(B_{N,0})&\le 1-\P\big(\xi_i<(6N)^{1/N}\ \forall i=1,\dots,N\big) \\ &\le1-\P\big(\xi_1<(6N)^{1/N}\big)^N=:q<1, \end{align*} since $1\in S$ and $(6N)^{1/N}>1$. So, \begin{equation} \P(A_N)=\prod_{k=0}^\infty \P(B_{N,k})\le \prod_{k=0}^\infty q=0, \end{equation} for each natural $N$. So, \begin{equation} P:=\P\Big( \exists N \in \mathbb{N} \text{ s.t. } \forall k \ge 0\ \prod_{i=1}^{N}\xi_{i+kN} > 6N \Big) =\P\Big(\bigcup_{N \in \mathbb{N}} A_N\Big)=0\ne1. \end{equation}

If now $1\notin S$, then $\P(\xi_i>c)=1$ for some real $c>1$ and hence almost surely $\prod_{i=1}^{N}\xi_{i+kN}>c^N > 6N$ for all $k$ and all large enough $N$. Hence, in this case $P=1$.

$\newcommand{\al}{\alpha} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\om}{\omega} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}} \newcommand{\tf}{\widetilde{f}}$

The answer depends on whether the support (say $S$) of the distribution of $\xi_1$ contains $1$ or not. If $1\in S$, then the answer is no; otherwise, yes.

Indeed, suppose first that $1\in S$. For each natural $N$, let \begin{equation} A_N:=\Big\{\forall k \ge 0\ \prod_{i=1}^{N}\xi_{i+kN} > 6N\Big\}=\bigcap_{k=0}^\infty B_{N,k}, \end{equation} where \begin{equation} B_{N,k}=\Big\{\prod_{i=1}^{N}\xi_{i+kN} > 6N\Big\}. \end{equation} The events $B_{N,0},B_{N,1},\dots$ are independent and for each $k$ \begin{align*} \P(B_{N,k})=\P(B_{N,0})&\le 1-\P\big(\xi_i<(6N)^{1/N}\ \forall i=1,\dots,N\big) \\ &=1-\P\big(\xi_1<(6N)^{1/N}\big)^N=:q<1, \end{align*} since $1\in S$ and $(6N)^{1/N}>1$. So, \begin{equation} \P(A_N)=\prod_{k=0}^\infty \P(B_{N,k})\le \prod_{k=0}^\infty q=0, \end{equation} for each natural $N$. So, \begin{equation} P:=\P\Big( \exists N \in \mathbb{N} \text{ s.t. } \forall k \ge 0\ \prod_{i=1}^{N}\xi_{i+kN} > 6N \Big) =\P\Big(\bigcup_{N \in \mathbb{N}} A_N\Big)=0\ne1. \end{equation}

If now $1\notin S$, then $\P(\xi_i>c)=1$ for some real $c>1$ and hence almost surely $\prod_{i=1}^{N}\xi_{i+kN}>c^N > 6N$ for all $k$ and all large enough $N$. Hence, in this case $P=1$.

$\newcommand{\al}{\alpha} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\om}{\omega} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}} \newcommand{\tf}{\widetilde{f}}$

The answer depends on whether the support (say $S$) of the distribution of $\xi_1$ contains $1$ or not. If $1\in S$, then the answer is no; otherwise, yes.

Indeed, suppose first that $1\in S$. For each natural $N$, let \begin{equation} A_N:=\{\forall k \ge 0, \prod_{i=1}^{N}\xi_{i+kN} > 6N\}=\bigcap_{k=0}^\infty B_{N,k}, \end{equation} where \begin{equation} B_{N,k}=\{\prod_{i=1}^{N}\xi_{i+kN} > 6N\}. \end{equation} The events $B_{N,0},B_{N,1},\dots$ are independent and for each $k$ \begin{multline*} \P(B_{N,k})=\P(B_{N,0})\le 1-\P(\xi_i<(6N)^{1/N}\ \forall i=1,\dots,N\} \\ \le1-\P(\xi_1<(6N)^{1/N})^N=:q<1, \end{multline*} since $1\in S$ and $(6N)^{1/N}>1$. So, \begin{equation} \P(A_N)=\prod_{k=0}^\infty \P(B_{N,k})=0, \end{equation} for each natural $N$. So, \begin{equation} P:=\P( \exists N \in \mathbb{N} \text{ s.t. } \forall k \ge 0 \prod_{i=1}^{N}\xi_{i+kN} > 6N ) =\P(\bigcup_N A_N)=0\ne1. \end{equation}

If now $1\notin S$, then $\P(\xi_i>c)=1$ for some real $c>1$ and hence $\prod_{i=1}^{N}\xi_{i+kN}>c^N > 6N$ for all $k$ and all large enough $N$. Hence, in this case $P=1$.

$\newcommand{\al}{\alpha} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\om}{\omega} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}} \newcommand{\tf}{\widetilde{f}}$

The answer depends on whether the support (say $S$) of the distribution of $\xi_1$ contains $1$ or not. If $1\in S$, then the answer is no; otherwise, yes.

Indeed, suppose first that $1\in S$. For each natural $N$, let \begin{equation} A_N:=\Big\{\forall k \ge 0\ \prod_{i=1}^{N}\xi_{i+kN} > 6N\Big\}=\bigcap_{k=0}^\infty B_{N,k}, \end{equation} where \begin{equation} B_{N,k}=\Big\{\prod_{i=1}^{N}\xi_{i+kN} > 6N\Big\}. \end{equation} The events $B_{N,0},B_{N,1},\dots$ are independent and for each $k$ \begin{align*} \P(B_{N,k})=\P(B_{N,0})&\le 1-\P\big(\xi_i<(6N)^{1/N}\ \forall i=1,\dots,N\big) \\ &\le1-\P\big(\xi_1<(6N)^{1/N}\big)^N=:q<1, \end{align*} since $1\in S$ and $(6N)^{1/N}>1$. So, \begin{equation} \P(A_N)=\prod_{k=0}^\infty \P(B_{N,k})\le \prod_{k=0}^\infty q=0, \end{equation} for each natural $N$. So, \begin{equation} P:=\P\Big( \exists N \in \mathbb{N} \text{ s.t. } \forall k \ge 0\ \prod_{i=1}^{N}\xi_{i+kN} > 6N \Big) =\P\Big(\bigcup_{N \in \mathbb{N}} A_N\Big)=0\ne1. \end{equation}

If now $1\notin S$, then $\P(\xi_i>c)=1$ for some real $c>1$ and hence almost surely $\prod_{i=1}^{N}\xi_{i+kN}>c^N > 6N$ for all $k$ and all large enough $N$. Hence, in this case $P=1$.