DFA Infiniteness Testing Why is it that the test for an infinite language on the DFA $M$ has an upper bound of $2n$?

That is to say, when testing for a string of length $k$, why does the theorem say:

$$n \leq k &lt; 2n$$

I understand the lower bound of $n$ IAW the Pumping Lemma of $|w| \geq n$. I'm just not sure why $2n$ was chosen and I think I'm over complicating it.
The upper bound comes also from the proof of the pumping lemma. If you have a DFA with $n$ states, you know that if it accepts a word in length $\ge n$ then the word passes through a circle in the states, thus you can repeat the circle as many times as you like and get an infinite language.

Now, for the upper bound, assume the DFA is infinite, and accepts a word of length $\ge2n$. Now, when you run the word on the DFA, and since the DFA has only $n$ states, there must be a cycle. That cycle must be of length $\le n$. You can now remove a number iterations from that circle such that you will get a word $w$ of length $n\le |w| &lt; 2n$.

The proof is even easier when you use the pumping length $n$ derived from the pumping lemma. Just show that for a word $w=xyz$ where $|w|\ge 2n$, since $|xy|\le n$ and $|y|&gt;0$ there exists an $i\in \mathbb{N}$ such that $n\le |xy^iz| &lt; 2n$.