Let $\left(a_n \in (0,2)\right)_{n\in\mathbb{N}}$; what does $\sum_{n \in \mathbb{N}} a_n \left( 2 - a_n \right) = +\infty$ mean in physical sense? I am reading a paper that states

Let $\left(a_n \in (0,2)\right)_{n \in \Bbb N}$ be sequence and satisfy
$$
\sum_{n\in\Bbb N}  a_n \left( 2 - a_n \right) = +\infty.
$$

I am sorry for asking probably trivial question. But for my learning,

What does the above condition signify (physical meaning sense would be appreciated)?
ADDED: Does this sequence converge or diverge?


NOTE: I am just not able to interpret the above condition. The sum of such sequence can not be $-\infty$ because every sequence is bounded, that is, $a_n \in (0,2)$. My thinking is that condition should have inequality but it has equality. Is this a way to show convergence (or divergence)? I am so confused.

 A: One interpretation is that the convergence/divergence of a series $\sum_n b_n$ with each $b_n \geq 0$ determines how quickly $b_n \to 0,$ if at all. Naturally you need $b_n \to 0$ for the series to converge, but if it doesn't happen quick enough (as noted in the comments) this may not hold true.
In that regard, we see that $a_n(2-a_n)$ is small if $a_n$ is close to $0$ or $2,$ so if we have convergence of the series
$$ \sum_{n \in \Bbb N} a_n(2-a_n) < \infty, $$
then this necessarily means we have the $a_n$'s concentrate at $0$ and/or $2.$ The fact that that series diverges, at least heuristically, tells us that this concentration either does not occur, or it doesn't happen very quickly.
Note this is a very vague interpretation of course, and I'm not sure if there's much else you can say in this regard.

Note that if $\sum_n a_n(2-a_n)$ diverges, then $\sum_n a_n$ does necessarily diverge. This is because $(2-a_n) \leq 2,$ so we have
$$ \sum_{n \in \Bbb N} a_n \geq \frac12 \sum_{n \in \Bbb N} a_n(2-a_n) = + \infty.$$
The converse is not true, intuitively because the sequence $(a_n)$ can concentrate near $2$ instead. For a concrete counterexample consider $a_n = 2 - \frac1{n^2}.$
