# 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!

## 1 Answer

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.

I don't think the author is trying to say anything about the boundedness or otherwise of the sequence itself, just to show that it exists.

For the second part of your question, any finite set of integers (or real numbers , for that matter) is clearly bounded below and above, so any unbounded set must be infinite.