What is the infimum of this set?

Question

I read the article on Infimum and supremum on wiki and came across this example. I am unable to understand Why the infimum of this set should be $-1$. I think it should be zero.

Can anyone help. $$ \inf\left\{ (-1)^n + \frac{1}{n}\bigg|\ n = 1,2,3,... \right\} = -1 $$

Answer 1

This is because, for $n=2k+1$ odd, we have $(-1)^n+\frac{1}{n}=-1+\frac{1}{2k+1}$, and for big $k$ those numbers can get arbitrarily close to $-1$. More rigorously:

• On one hand, we have $(-1)^n+\frac{1}{n}\ge -1$ for all $n$, so $-1$ is a lower bound of the whole set.
• On the other hand, this is the highest lower bound: if you pick anything higher than that, say $-1+\epsilon$ for some $\epsilon\gt 0$, there will be elements of this set "below" that: whenever $2k+1\gt\frac{1}{\epsilon}$, we will have $-1+\frac{1}{2k+1}\lt-1+\epsilon$.

Thus $-1$ is the highest lower bound, i.e. infimum of this set.

Answer 2

$$\begin{align}S&=\left\{(-1)^n+\dfrac1n\bigg|n=1,2,3\cdots\right\}\\&=\begin{cases}-1+\dfrac1n,&n=2k+1\\1+\dfrac1n,&n=2k\end{cases}\end{align}$$

Taking limits as $n\to\infty$ $$\begin{align}S&=\begin{cases}(-1,0],&n=2k+1\\ \left(1,\dfrac32\right]&n=2k\end{cases}\end{align}$$

Hence $$\begin{align}\sup(S)&=\dfrac32\\\inf(S)&=-1\end{align}$$