Sup, Inf, max and min of a given set For a given set $$M_1 =  \left\lbrace \frac{2-(-1)^{n}}{n}; n \in \mathbb{N} \right\rbrace. $$
Determine if it is bounded? Determine inf and sup, and if they exist max and min.
My first concern is if $n = 0$, we have an undefined value in the set, what does this mean for min, max, inf and sup?
If I continue without this concern I determine $\max = 3 $, when $ n=1 $ since the given number will be decreasing as $n$ increases.
My guess for $\min$ is that it does not exist, since the series will be approaching 0 as $n \to \infty $.
Should $\inf = 0$ suffice as an answer? How can I motivate this?
I am unsure as how to choose $\sup$ since it would need to be larger than 3, but can I choose any number larger? How can I motivate it?
Please take into account the level I am at, I am taking my second semester of maths at university and we have just started with the definition of limits and sup/inf.
 A: 
My first concern is if $n = 0$, we have an undefined value in the set,
  what does this mean for min, max, inf and sup?

While there is no consensus, it is fairly usual to write $\mathbb N$ to mean the set $\{1,2,3,4\dots\}$, in other words, $0\notin\mathbb N$.


If I continue without this concern I determine $\max = 3 $, when $ n=1 $ since the given number will be decreasing as $n$ increases.

While $\max=3$ is the correct answer, your reasoning is incorrect. The given number is not decreasing as $n$ increases. When you go from $n=2$ to $n=3$, the number does not decrease. It goes from $\frac{2-1}{2}=\frac12$ to $\frac{2+1}{3}=1$, so it increases!


My guess for $\min$ is that it does not exist, since the series will be approaching 0 as $n \to \infty $.

Again, your guess is correct, but the reasoning is not. True, the series approaches $0$ as $n\to \infty$, but that is not enough to conclude that the minimum does not exist. For example, the series $$0, 1, \frac12, \frac13,\frac14,\frac15,\dots$$ also approaches $0$ as $n\to\infty$, but the set of its elements does have a minimum!


Should $\inf = 0$ suffice as an answer? How can I motivate this?

You can prove that $0$ is the infimum by proving


*

*That $0$ is a lower bound

*That for every $\epsilon > 0$, the value $0+\epsilon$ is not a lower bound.




I am unsure as how to choose $\sup$ since it would need to be larger than 3, but can I choose any number larger? How can I motivate it?

If a set has a maximum element, then that element is always also the supremum. This is a very simple claim to prove rigorously, from the definition of supremum (it is great practice if you have the time!).
A: You have got most of it right. When a set has  a maximum the maximum and supremum coincide. So the supremum is $3$ and the infimum is $0$. As far as your first question is concerned, $\mathbb N$ is the usual notation for $\{1,2,...\}$ and $0$ is not included.  
