Q:

convergence of a series $a_1 + a_1 a_2 + a_1 a_2 a_3 +\cdots$

Suppose all $a_n$ are real numbers and $\lim_{n\to\infty} a_n$ exists.
What is the condition for the convergence( or divergence ) of the series
$$ a_1 + a_1 a_2 + a_1 a_2 a_3 +\cdots $$
I can prove that $ \lim_{n\to\infty} |a_n| < 1 $ ( or > 1 ) guarantees
absolute convergence ( or divergence ).
What if $ \lim_{n\to\infty} a_n = 1 \text{ and } a_n < 1 \text{ for all } n $ ?

A:

What if $\lim_{n\to\infty}a_n=1$ and $a_n<1$ for all $n$?

Then the series may or may not converge. A necessary criterion for the convergence of the series is that the sequence of products
$$p_n = \prod_{k = 1}^n a_k$$
converges to $0$.
If the $a_n$ converge to $1$ fast enough, say $a_n = 1 - \frac{1}{2^n}$ ($\sum \lvert 1 - a_n\rvert < \infty$ is sufficient, if no $a_n = 0$), the product converges to a nonzero value, and hence the series diverges.
If the convergence of $a_n \to 1$ is slow enough ($a_n = 1 - \frac{1}{\sqrt{n+1}}$ is slow enough), the product converges to $0$ fast enough for the series to converge.
Let $a_n = 1 - u_n$, with $0 < u_n < 1$ and $u_n \to 0$. Without loss of generality, assume $u_n < \frac14$.
Then $\log p_n = \sum\limits_{k = 1}^n \log (1 - u_k)$. Since for $0 < x < \frac14$, we have $-\frac32 x < \log (1-x) < -x$, we find
$$ -\frac32 \sum_{k=1}^n u_k < \log p_n < -\sum_{k=1}^n u_k,$$
and thus can deduce that if $\sum u_k < \infty$, then $\lim\limits_{n\to\infty}p_n > 0$, so the series does not converge.
On the other hand, if $\exists C > 1$ with $\sum\limits_{k = 1}^n u_k \geqslant C\cdot \log n$, then $p_n < \exp (-C \cdot\log n) = \frac{1}{n^C}$, and the series converges.

