How do I read this inequality $0\leq k < 0$

Question

In this answer the summation index is indicated as $0\leq k < 0$.

I read this as,

k is less than zero
k equals zero
k is greater than zero

What is the value of $k$ ? Can $k$ be both less than and greater than zero? It seems that there is a simple convention that I'm not aware of. Can you please explain?

Also, in the same answer, there is the word "anything" inside the summation sign. Is "anything" a mathematical term? Probably it is meant "any function of $k$ goes here"?

Answer 1

Your point about "can $k$ be both gretaer than and less or equal to zero" hits right on the mark here: the point is there there is no such $k$! This means that we have an empty sum: as written in that answer, $$\sum_{0 \le k < 0} f(k) = 0.$$ "anything" stands, as you say, for any arbitrary function of $k$.

Answer 2

The sum $\sum_{0\le k<0}2^k$ asks us to add up terms $2^k$ for integers $k$ such that $0\le k < 0$.

Since there are no such $k$, $\sum_{0\le k<0}2^k=0$.

Observe $\{k \in \mathbb Z : 0\le k < 0\}$ is the empty set. So, we call $\sum_{0\le k<0}2^k$ an empty sum.

Answer 3

In my opinion, summation over an empty set can’t be defined using the standard meaning for summation, as you aptly describe ‘the summation that did not happen’ in one of your comments. However, due to its use in several mathematical formulae, it satisfies many situations to set such a summation to be equal to the additive identity ($0$). Therefore, I think it is a case of expansion of the originally intended definition to better suit many situations.

To put it another way, for two disjoint sets $A$ and $B$, if you define summation over an empty set to be $0$, the relation $$\sum_{x\in A\cup B}f(x)=\sum_{x \in A}f(x)+\sum_{x \in B}f(x)$$will still hold even when $A$ or $B$ or both is/are the empty set $\phi$

So defining summation over an empty set to be equal to $0$ is merely a choice, albeit a very good one, which is accepted by most; as it extends many already present formulae to a higher domain.