I can prove the following proposition:
(*) let $a_n>1$ for all $N\le n.$ $\prod_{n=1}^\infty a_n$ converges if and only if $\sum_{n=1}^\infty (a_n-1)$ converges.
Proof:
let $b_n=a_n-1$. since, for all $N\le n$, $a_n>1$, we have $b_n>0$.
Note $\prod_{n=1}^\infty a_n$ converges iff $\sum_{n=1}^\infty ln(a_n)$ converges iff $\sum_{n=1}^\infty ln(1+b_n)$ converges.
Consider $lim_{t\to \infty}\frac{ln(1+t)}{t}=1$, which implies $\exists\delta>0$ s.t $\forall t\in (-\delta,\delta), |\frac{ln(1+t)}{t}-1|<\frac{1}{2},$ therefore, $\forall t\in (0,\delta), \frac{t}{2}<ln(1+t)<\frac{3t}{2}$.
Note that no matter which side we start, $lim$ $b_n=0$.
if $lim$ $b_n=0$, $\exists n_0\in \Bbb N$ such taht $N<n_0$ and $\forall N\le n$, $0<b_n<\delta$, therefore, $\frac{b_n}{2}<ln(1+b_n)<\frac{3b_n}{2} $ and $\forall n_0<m, 0<\frac{1}{2}\sum_{n=n_0}^mb_n<\sum_{n=n_0}^mln(1+b_n)<\frac{3}{2}\sum_{n=n_0}^mb_n$
Therefore, by comparison test, the result follows.
But I got stuck to prove another similar proposition, which reduce the requirement of $1<a_n$ to $0<a_n$ and require the convergence of $\sum_{n=1}^\infty b_n^2$
(*) let $a_n>0$ for all $N\le n.$ if $\sum_{n=1}^\infty (a_n-1) ^2$ converges , then $\prod_{n=1}^\infty a_n$ converges if and only if $\sum_{n=1}^\infty (a_n-1)$ converges.
any idea and suggestion will be appreciated.