Implication from the convergence of an infinite product Let $(a_k)_{k \in \mathbb{N}}$ be a sequence of non-negative real numbers such that
$$
\prod_{k=1}^{\infty} a_k \in (0,1].
$$
Based on the hint in the comments of this post, it is possible to show that for any $m\in\mathbb{N}$
$$
\lim_{ n \rightarrow \infty}\prod_{k=n+1}^{n+m} a_k =1. \tag{1}
$$
One can write
$$
\prod_{k=1}^{n+m} a_k = \prod_{k = 1}^{n} a_k \prod_{k = n + 1}^{n+m} a_k,
$$
hence
$$
\prod_{k = n + 1}^{n+m} a_k = \frac{\prod_{k=1}^{n+m} a_k}{\prod_{k = 1}^{n} a_k},
$$
and taking the limit as $n \rightarrow \infty$ yields the result.

Now, by definition, $(1)$ means that for every $m \in \mathbb{N}$ and for every $\varepsilon > 0$ there exists $N(m, \varepsilon) \in \mathbb{N}$ such that
$$
\left| \prod_{k=n+1}^{n+m} a_k - 1 \right| < \varepsilon \quad \forall n \geq N(m, \varepsilon). \tag{2}
$$

If we further assume that $\prod_{k=1}^{n} a_k \leq 1$ $\forall n \in \mathbb{N}$, is it possible to choose $N$ independent of $m$ in $(2)$.

 A: First note that the exponential function is continuous, and for all $\epsilon >0$ there exists $\delta(\epsilon) > 0$ such that $|\exp(x) - \exp(0)| =|\exp(x) - 1|< \epsilon$ when $|x - 0| < \delta(\epsilon)$.
It is given that $a_k \geqslant 0$ for all $k$ and the infinite product $\prod_{k=1}^\infty a_k$ converges to a value $P \in (0,1]$.  Hence, we must have $a_k > 0$ for all $k$. This avoids any problems with taking logarithms.
Consider  the partial products $P_n = \prod_{k=1}^na_k$. Since $P_n \to P$ it follows that   $\log P_n\to \log P$ as $n \to \infty$, by continuity of the logarithmic function.  Hence, $(\log P_n)$ must be a Cauchy sequence, and given $\delta(\epsilon)$ there exists $N(\epsilon) \in \mathbb{N}$ such that for all $n > N(\epsilon)$ and all $m \in \mathbb{N}$, we have
$$|\log P_{m+n} - \log P_n| < \delta(\epsilon),$$
(See a discussion of  Cauchy sequence in any analysis textbook to confirm that $N(\epsilon)$ is indepedent of $m$).
Hence,
$$\left| \log \prod_{k=n+1}^{n+m} a_k - \log (1)\right|= \left| \log \prod_{k=n+1}^{n+m} a_k \right|= \left|\log \frac{P_{n+m}}{P_n} \right|= |\log P_{n+m} - \log P_n| < \delta(\epsilon),$$
and by the continuity of the exponential function,
$$\tag{*}\left|\prod_{k=n+1}^{n+m} a_k -1\right| = \left|\exp\left(\log\prod_{k=n+1}^{n+m} a_k   \right) - \exp(\log 1) \right| < \epsilon,$$
where (*) holds for all $n > N(\epsilon)$ and all $m \in \mathbb{N}$.
