Determine the $\sup(A)$ if $A = \{ \frac{(-1)^n}{n}: n \in \mathbb{N} \}$ and prove your result.

Question

It is clear that $\frac{1}{2} \geq \frac{(-1)^n}{n}, \forall n \in \mathbb{N} \implies \frac{1}{2}$ is an upper bound of $A$.

My intention is to make use of the lemma,

Assume $s \in \mathbb{R}$ is an upper bound for a set $A \subset \mathbb{R}$. Then, $s = sup(A)$ iff, $\forall \epsilon > 0$, $\exists a \in A$ satisfying $s - \epsilon < a$.

Let $\epsilon > 0$ be arbitrary. There are two cases,

Case 1: $\epsilon = \frac{1}{2}$

Choose $n = 2$ (Note that this particular $n$ works $\frac{1}{2} > \frac{1}{2} - \frac{1}{2} = 0$)

Case 2: $\epsilon \neq \frac{1}{2}$

Choose some $n$ satisfying $n < \frac{1}{\sqrt{\epsilon^2 - \epsilon + \frac{1}{4}}} \implies \frac{1}{n^2} > \epsilon^2 - \epsilon + \frac{1}{4} \implies \frac{(-1)^n}{n} > \frac{1}{2} - \epsilon$

QED.

I mostly seek for a solution that makes use of that lemma, but I am willing to be open to any solutions that you come up with.

Answer 1

There is no need to separate the cases. No matter what $\epsilon$ you take, it is always true that you can choose $\frac12\in A$ (i.e., the element you get if you choose $n=2$).

Because $\frac12\in A$, it is clear that whatever $\epsilon>0$ you are given, you have

$$\frac12 > \frac12-\epsilon$$

and this is already enough.

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In fact, this is a good starting point to prove a more general result:

If $m$ is an upper bound for $A$, and $m\in A$, then $m=\sup(A)$.

You can prove this statement more or less exactly from your lemma.

Answer 2

Since $\frac12$ is an upper bound of $A$ and since $\frac12\in A$, $\frac12=\sup A$.

If you want to use the lemma, you can do it this way: if $\varepsilon>0$, then, $\frac12-\varepsilon<\frac12$. Therefore, $\frac12=\sup A$.