Show that the $\sup S=1/2$ where $S= \{(-1)^n/n : n \in \mathbb{N} \}$

Question

Show that the sup(S) where $S = \{\frac{(-1)^{n}}{n} : n \in \mathbb{N} \}$ is sup(S) = $\frac{1}{2}$

Attempt:

In order to show $\frac{1}{2}$ is the supremum I must show:

$$\forall \epsilon > 0, \exists s\in S \ such\ that \\ \frac{1}{2}-\epsilon < s < \frac{1}{2}$$

So from this statement what I gather is I have to find an 's' that would be a function of $\epsilon$.

My question is how do I find that s?

I think what I should do is given an element $s \in S$ is of the form $$\frac{(-1)^{n}}{n}$$

Just solve $$\frac{1}{2} - \epsilon = \frac{(-1)^{n}}{n}$$

But then in this expression there is no s. So am I solving for s or $\epsilon$?

Answer 1

You are mistaken in your characterization of the supremum. To show $\sup S=\frac12$, you must show two things:

1. For all $s\in S$, $s\le \frac12$. [This means $\frac12$ is an upper bound]

2. For all $\epsilon >0$, there exists $s\in S$ so $s>\frac12-\epsilon$. [This means there can be no smaller upper bounds].

Here are the proofs:

1. If $n$ is odd, then $(-1)^n/n<0<\frac12$. If $n$ is even, so $n\ge 2$, then $(-1)^n/n=1/n\le \frac12$.

2. Choose $s = \frac12\in S$. Then $s=\frac12>\frac12-\epsilon$.

Answer 2

The supremum of a set is, first of all, an upper bound of the set. It is very easy to check that for any $s \in S$ it holds that $1/2 \geq s$. Set $s^* = 1/2$. We will show that $s^*$ using the definition you mentioned.

We already proved that $s^* \geq s$ for any $s \in S$. Now let $\epsilon >0$. Then we know that $1/2 \in S$ and moreover, $s^* - \epsilon < 1/2$. Therefore we found an element of the set that is greater than $s^* -\epsilon$ for any $\epsilon >0$. Thus, $s^*=1/2$ is the supremum.