# Question about properly defined sequence

I was reading through Fitzpatrick's Advanced Calculus and in chapter 2.1, they list an example of a sequence as follows.

Given a natural number $n$ define the sequence $a_{n}$ to be the sequence in which the $n^{th}$ term is the largest integer that is less than or equal to $\sqrt{n^{3}}$.

For example, the firts four terms in the series are

1, 2, 3, 8

By a previous theorem proved in the book, any nonempty set of integers that is bounded above has a largest member.

Then, the book goes on to say that because there is always a largest integer less than or equal to $\sqrt{n^{3}}$, that the sequence $\{a_{n}\}$ is properly defined.

My first question is, what does a properly defined sequence entail? They gloss over this definition in the book.

My second question is, do we only care that given any n, that we have a set of terms up to that nth term in which the set is bounded above? I was a little confused because I didn't see how the sequence $\{a_{n}\}$ could be bounded above, as I can always pick a larger n to get a larger term in the sequence, and it seems pretty clear that the sequence is divergent, though I haven't tested this yet. The book has mentioned convergence or divergence tests yet so I didn't think I would need them to understand this part.

Thanks!

By "properly defined" the book seems to mean that for each $n$ there is a single integer $a_n$ which fits the definition. So you have a function $a_:\mathbb N\to \mathbb Z$ for which $n\to a_n$, which is what is meant by a sequence of integers.
The fact that $a_n$ is well-defined comes because $\sqrt {n^3}$ is an upper bound for $a_n$ and the set $\{z\in \mathbb Z: z\lt \sqrt {n^3}\}$ is non-empty (take $z=0$). So you can use the previous theorem to show that $a_n$ exists and is unique.