What's the supremum of $\left\{(-1)^n(1-\frac{1}{n^2}): n \in \mathbb{N}\right\}$?

Question

I wish to find $\displaystyle \sup(A)$ for $\displaystyle A = \left\{(-1)^n\left(1-\frac{1}{n^2}\right): n \in \mathbb{N}\right\}$.

Definition: $u$ is the supremum of $A$ if for any $\epsilon >0$ we have $x \in A$ implies $x+\epsilon > u$.

First we find an upper bound: by triangle inequality $\displaystyle (-1)^n \left(1-\frac{1}{n^2}\right) \le 1+\frac{1}{n^2} \to 1. $

So $1$ is an upper bound. We'll prove that it's the least upper bound by contradiction. Suppose for every $\epsilon >0$ we have $x \in A$ implies $x+\epsilon \le 1$. That's, we have $\epsilon \le 1-x$ . But $x$, being in A which is bounded by $1$, is less than or equal to $1$. So $\epsilon \le 1-x \le 0$. Hence $\epsilon \le 0$, contradicting that $\epsilon >0$.

Is this right? I'm not terribly confident because I wasn't able to fix an epsilon. Thanks in advance.

Answer 1

You don't have to “fix an epsilon”.

First you make the conjecture that $1$ is the supremum. It clearly is an upper bound, because $1>(-1)^n(1-1/n^2)$.

In order to show it is the least upper bound, take $\varepsilon>0$. You need to find $n=2k$ (why does it suffice to be even?) such that $$ 1-\frac{1}{4k^2}>1-\varepsilon $$ If $\varepsilon\ge1$ there is no problem, as any $k$ is good. Suppose $0<\varepsilon<1$. Then the inequality becomes $$ k>\frac{1}{2\sqrt{\varepsilon}} $$ and surely such a $k$ exists.

Therefore, for every $\varepsilon>0$, there exists $n$ such that $$ (-1)^n\left(1-\frac{1}{n^2}\right)>1-\varepsilon $$ and we're done.

Answer 2

$1$ is an upper bound, right. It is also the least upper bound, i.e., $1=\sup A$.

Indeed, suppose by contradiction that $x:=\sup A$ is smaller than $1$. Then there exists a large even $n$ such that $$ x<(-1)^n \left(1-\frac{1}{n^2}\right)<1. $$

Answer 3

Let me rephrase a bit.

$1$ is an upper bound.

Assume:

$b\lt 1$ is an upper bound.

Let $\epsilon = (1/2)(1-b)>0$.

Archimedes:

There is a $m \in \mathbb{N}$ such that:

$m \gt 1/\epsilon.$

Then for $(2n)^2 >m >1/\epsilon $:

$\epsilon >1/m>1/(2n)^2$, or

$\epsilon = (1/2)(1-b) >1/(2n)^2$ .

We have $1-\epsilon < 1- 1/(2n)^2:$

$1-(1/2)(1-b) = 1/2 +(1/2)b < 1-1/(2n)^2$ , or

$b = b/2 +b/2 < 1/2+b/2$

$< 1-1/(2n)^2 <1,$

a contradiction,

hence $\sup(A)=1.$