# Infimum of a set (II)

I came across another question about infimums of sets.

Find $\inf \{(-1)^{n} + 1/n: n \in \mathbb{Z}^{+} \}$.

Heuristaically, I think the infimum is $-1$ since, for large $n$, the second term goes to $0$ and the first term is at least $-1$. Let $A = \{(-1)^{n} + 1/n: n \in \mathbb{Z}^{+} \}$. I know that $-1$ is a lower bound for $A$ (would I have to justify this?). So $\alpha = \inf A$ exists. Thus $-1 \leq \alpha$. Now I need to show that $\alpha \leq -1$. Suppose for contradiction that $\alpha > -1$. Then I would need to find a $\beta$ such that $\beta > \alpha > -1$ where $\beta$ is also a lower bound?

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If $\alpha \gt -1$, you should be able to find an element $\beta \lt \alpha$ of your set, so $\alpha$ is not a lower bound. – Ross Millikan Jun 21 '11 at 19:06

The definition of " $\alpha$ is the infiumum of $A$" is:
• $\alpha \leq a$ for all $a\in A$; and
• If $\beta\gt \alpha$, then $\beta$ is not a lower bound for $A$; that is, if $\beta\gt \alpha$, then there exists $a\in A$ such that $\alpha\leq a\lt \beta$.
So, you know that for $A=\{(-1)^n + 1/n\mid n\in\mathbb{Z}^+\}$, $\alpha=-1$ is a lower bound. You are conjecturing that it is, in fact, the infimum. Finding a strictly larger $\beta$ that is also a lower bound would indeed show that it is not the infimum, but you'll have a hard time doing that directly.
Instead, try showing that if $\beta\gt -1$, then there exists $a\in A$ with $-1\leq a\lt \beta$. That will show that no number greater than $-1$ is a lower bound.