Let $a_1$ be linearly independent to $a_2$ over $\mathbb{Q}.$ For $n\geq 3,$ let $ a_n = \vert a_{n-1} - a_{n-2} \vert.$ Does $\sum_n a_n\ $ converge? Let $a_1$ be linearly independent to $a_2$ over the rational numbers. For $n\geq 3,\ $ let $ a_n = \vert a_{n-1} - a_{n-2} \vert.$ Does $\sum_n a_n\ $ converge?
For example, let $a_1 = 1,\ a_2 = \ln 2=0.693\ldots\ .\ $ Then, $\ a_3 = \vert a_2 - a_1 \vert = 0.306\ldots,\ \ a_4 = \vert a_3 - a_2 \vert = 0.386\ldots,\ \ a_5 = \vert a_4 - a_3 \vert = 0.0794\ldots\ .$ I am not sure how to judge how quickly this converges to $0.$
 A: I'll show there are cases where the sum $\sum a_n$ diverges.
Let's assume $a_1=u$, $a_2=v$ with $u>v$ and $\lfloor u/v\rfloor=2k$. Then, the sequence becomes
$$
u,v,u-v,u-2v,v,u-3v,u-4v,v,\ldots,u-2kv,v,(2k+1)v-u,\ldots
$$
where the term $(2k+1)v-u<v$ is the first that breaks the pattern as $(2k+1)v>u$.
Now, note that if we sum the terms $u,v,\ldots,u-2kv$, the sum is at least $ku$ as the terms on the form $u-iv$ may be paired up into $k$ pairs, each with sum $>u$ (ignoring the $v$ terms and one of the $u-iv$ terms). The remainder of the sequence now starts with $v,(2k+1)v-u$ on which we may repeat the above argument.
If we let $b_0=u, b_1=v$, we seek to get $2k_m=\lfloor b_{m-1}/b_m\rfloor$ and $b_{m+1}=(2k_m+1)b_m-b_{m-1}$. If we succeed in this, the sequence will become
$$
\ldots,b_{m-1},b_m,b_{m-1}-b_m,\ldots,b_{m-1}-2k_mb_m,b_m,b_{m+1}\ldots
$$
and the sum $\sum a_n > \sum k_mb_{m-1}$.
The relation in $b_m$ may be rewritten
$$
\frac{b_{m}}{b_{m-1}}
= \frac{1}{2k_m+1-\frac{b_{m+1}}{b_{m}}}
$$
which leads to the continuous fraction
$$
\frac{b_1}{b_0}
= \frac{1}{2k_1+1-\frac{1}{2k_2+1-\frac{1}{\ddots}}}.
$$
Let $r_m=b_m/b_{m-1}$. The continued fraction is defined by $r_m=1/(2k_m+1-r_{m+1})$ letting $r_N=0$, and letting $N\rightarrow\infty$. Since $0\le r_{m+1}<1$ implies $\frac1{2k_m+1}\le r_m\le\frac1{2k_m}$, this will also apply to the limit which defined the continued fraction.
Since $\sum a_n>\sum k_m b_{m-1}=\sum k_m r_{m-1}r_{m-2}\cdots r_1b_0$, we can make the sum diverge by forcing $k_m$ to be sufficiently large. Eg, if we let $k_m>(2k_{m-1}+1)\cdots(2k_1+1)$, this ensures $k_mr_{m-1}\cdots r_1>1$, making the sum diverge.
A: I was trying to figure out how to determine whether it converges given a specific $a_1, a_2$, and how to find out the sum when it converges. I didn't manage to fully solve it, but here is some progress that someone else might be able to build off.

WLOG, let $a_1=1$ and $a_2=d$ (you can assume this since the sequence terms are linearly proportional with $a_1$ when $a_2$ is scaled appropriately) and let $f(d)$ be the sum of $a_n$. If $d < 0$, the sequence goes $$1, d, 1-d, 1-2d, -d, 1-d, 1, -d, \cdots$$
so $f(d)=1+d+1-d+1-2d-d+1-d+f(-d)\to f(d)=4-4d+f(-d)$ for $d < 0$.
If $d>1$, then the sequence goes $$1, d, d-1, 1, |2-d|, \cdots$$
so $f(d)=1+d+d-1+f(|2-d|)\to f(d)=2d+f(|2-d|)$ for $d > 1$. Let $k = \left[\frac{d}{2}\right]$, where $[x]$ denotes the nearest integer to $x$, so that $2k$ is the even integer closest to $d$. Then unwrapping this recursion using this yields that $$f(d)=2kd-2k(k-1)+f(|d-2k|)\tag 1$$ for $d > 1$, so that $|d-2k|\in(0,1)$ (this actually holds for $0 < d < 1$ too, but that would just say that $f(d)=f(|d|)$ which is trivial).
That leaves $0 < d < 1$. Let $k = \left\lfloor\frac{1}{d}\right\rfloor$. There are two cases here depending on the parity of $k$. If $k$ is odd, the sequence would go $$1, d, 1-d, 1-2d, d, 1-3d, 1-4d, d, 1-5d, \cdots, d, 1-kd, (k+1)d-1$$ where every third term is a $d$. If $k$ is even, the sequence would go $$1, d, 1-d, 1-2d, d, 1-3d, 1-4d, d, 1-5d, \cdots, 1-(k-1)d, 1-kd, d, (k+1)d-1$$ where again, every third term is a $d$.
With $k$ even: $$f(d)=\frac{k}{2}d+\sum_{r=0}^k(1-rd)+d\cdot f\left(k+1-\frac{1}{d}\right)$$ which simplifies to $$f(d)=-\frac{k^2}{2}\cdot d+k+1+d\cdot f\left(k+1-\frac{1}{d}\right)$$ (this actually holds for $k = 0\iff d>1$  as well)
With $k$ odd: $$f(d)=\frac{k+1}{2}d+\sum_{r=0}^{k-1}(1-rd)+d\cdot f\left(-k+\frac{1}{d}\right)$$ which simplifies to $$f(d)=\frac{2-(k-1)^2}{2}d+k+d\cdot f\left(-k+\frac{1}{d}\right)$$
Let $d\in(0,1)$ be represented as $[k_1, k_2, \cdots]$ and let $d_n$ be $[k_n, k_{n+1}, \cdots]$. Generate a sequence $k_n$ such that $0<d_n<1$ and if $k_n$ is odd, $d_n=\frac{1}{k_n+d_{n+1}}$, and if $k_n$ is even, $d_n=\frac{1}{k_n+1-d_{n+1}}$. This is similar to the continued fraction representation, but not exactly. As an example, $\frac{1}{\sqrt{2}}=\frac{1}{1+\frac{1}{2+1-\frac{1}{1+\frac{1}{ \sqrt{2}}}}}$, so $\frac{1}{\sqrt{2}}=[1,2,1,1,2,1,\cdots]$
Then if $k_n$ is odd, $$f(d_n)=\frac{2-(k_n-1)^2}{2}d_n+k_n+d_n\cdot f(d_{n+1})$$
and if $k_n$ is even, $$f(d_n)=-\frac{k_n^2}{2}\cdot d_n+k_n+1+d_n\cdot f(d_{n+1})$$
In order for $f(d)$ to exist, the limit as you apply this infinite times needs to exist.
If $d\in(0,1)$ has a periodic representation, then I believe that means that $f(d)$ exists. With $d=\frac{1}{\sqrt{2}}$, you'd get that $f(d)=2+\frac{7}{2\sqrt{2}}=2+\frac{7}{2}d$, and for $d=\frac{-1+\sqrt{5}}{2}$, you'd get that $f(d)=2+\sqrt{5}=3+2d$
It might be helpful to note that
$$\left|f(d_n)-1-\frac{1}{2d_n}-d_n\cdot f(d_{n+1})\right|\le \frac{d_n}{2}$$
This was obtained by setting bounds from the cases of $k_n$ being even and odd.
A: Using the linear independence, you can show easily that $a_n$ is never zero. Indeed, if $a_{n+2}=0$ for some $n$, then $a_{n+1} = a_n$ which also implies $a_{n-1}=0$. In the end, you would get $a_1a_2a_3=0$ which is impossible.
Using this fact $a_{n+2}=|a_{n+1}-a_n|< a_{n+1}$. That follows from $a_n>0$ (for $n>2$ in the case where $a_2<0$).
Thus, the sequence is decreasing and $a_{n+2}=a_n-a_{n+1}$ for $n>2$ so that it converges to zero. For the sum of the terms, you get a cancellation because
\begin{align*}
    \sum_{n=3}^Na_{n+2} = \sum_{n=3}^Na_{n}
-a_{n+1} = a_3 - a_{N+1} \to a_3.\end{align*}
