Can we conclude that the a sequence $a_n$ such that $ |a_1| \lt |a_2 -a_1| \lt |a_3 -a_2| \lt |a_4 - a_3| \dots$, and $a_1 \neq 0$ is increasing? We have an infinite sequence
$$
a_1, a_2 , a_3 \cdots 
$$
And it is given that
$$
|a_1| \lt |a_2 -a_1| \lt |a_3 -a_2| \lt |a_4 - a_3| \cdots \\
a_1 \neq 0
$$
(that is the difference between the subsequent terms are increasing and first term cannot be zero)
Can we conclude that the absolute values of subsequent terms are increasing? That is can we conclude
$$
|a_1| \lt |a_2| \lt |a_3| \lt |a_4| \cdots
$$
Playing around with the inequalities given in the question can give us the information that the alternate terms are increasing (in absolute/numerical value, leaving $a_1$ aside, that is, not comparing $a_1$ with any terms but just caring that it is not zero) but not the consecutive terms. So, I think we cannot conclude that the consecutive terms are numerically increasing.
An  explanatory answer is sought.
 A: One counterexample suffices, and you can produce one with just three terms. If you want to go a little further and show that there need not even be a point beyond which the terms increase in absolute value, you have to work a little harder, but not much. For instance, let $a_1=1$, and in general let
$$a_{n+1}=\begin{cases}
a_n-n,&\text{if }n\text{ is odd}\\
a_n+n,&\text{if }n\text{ is even,}
\end{cases}$$
so that you get the sequence $1,0,2,-1,3,\ldots\;$; it’s not hard to show by induction that in that case $a_{2n-1}=n$ and $a_{2n}=1-n$ for all $n\in\Bbb Z^+$. Evidently $|a_{n+1}-a_n|=n$ for $n\in\Bbb Z^+$, but $|a_{2n}|=n-1<n=|a_{2n-1}|$ for $n\in\Bbb Z^+$.
A: Consider the sequence $b_n:=c_n(1-\tfrac{1}{n})$ where $c_n\in\{+1,-1\}$, and the sequence $a_n:=\sum_{i=1}^nb_i$.
Then $|a_{n+1}-a_n|=|b_{n+1}|=1-\tfrac{1}{n+1}$ is increasing, yet because of the random choice of $c_i$ there is no telling whether $a_n$ is increasing or decreasing. Here is an example generated by a random choice of $c_n$.

