What is the difference between a supremum and maximum; and also between the infimum and minimum? What is the difference between a supremum and maximum; and similarly the infimum and minimum? Also, how does one tell if they exist?
Here is an example:

$$x_n = \frac{n}{2n-1}$$
Determine whether the maximum, the minimum, the supremum, and the
  infimum of the sequence $x_n$ n=1 to n=+∞

My understanding: 


*

*The limit is 1/2. 

*$x_n$ is decreasing. 

*The supremum exists as n goes to +∞. 

*The infimum does not exist as limited by n=1, minimum = 1/2. 

*The maximum and supremum exists, both equal to 1. 

 A: My answer is conceptual rather than an exact solution to your problem. I believe you will solve the problem after you understand the concepts.
Let's start with the definition of an upper bound. $M$ is an upper bound of the set $A$ if every element of $A$ is less than $M$. And the minimum of these $M$'s is called the supremum of $A$, denoted by $\sup A$. That is to say, if $M$ is an upper bound of $A$ and there is not an upper bound of $A$ smaller than $M$, then $\sup A = M$. And there is a fundamental property of real numbers called the least upper bound axiom which states the following:
A set $X$ has a supremum iff $X$ is non-empty and has an upper bound.
You can play with these definitions (use terms like lower bound instead of upper bound) to obtain the cases for the infimum.
And as others indicated, the difference between the supremum and the maximum is that maximum of a set must be contained by the set but supremum of a set may not be contained by the set. When the first case is satisfied, that is, if a set has a maximum, then the maximum is equal to the supremum as you can show easily using the definitions above. For example, if $S = [1,5]$ then $\sup S = \max S = 5$. But if $S = [1,5)$, then $\sup S = 5$ and $S$ has no maximum.
Therefore, one can easily show, for your example, that the set  $S =\left\{ {x_n \:\:\: |\:\:\: n \ge 1} \right\}$ has a lower bound (eg. $0$ is a lower bound for the set $S$), and is non-empty ($1 \in S$). Therefore, by the greatest lower bound property (which you should have obtained by playing the definitions above) the set has an infimum.
Remark: For some sets by using induction or other simple methods it is easy to show that whether the set has a supremum or not. But try to show that the set consisting of the numbers $$\bigg(1+ \frac{1}{n}\bigg)^n$$ where $n\ge 1$ has a supremum. You will see that it is not that easy.
A: The difference between supremum and maximum is that for bounded, infinite sets, the maximum may not exist, but the supremum always does. Consider the set $(0,1)$. Does this set contain a largest element? The answer is no because for any $x \in (0,1)$, $\tfrac{x+1}{2}$ is also in $(0,1)$ and $x < \tfrac{x+1}{2}$. That is, for any $x \in (0,1)$, we can find an element in $(0,1)$ which is larger than $x$. Thus there is no maximal element of $(0,1)$. By contrast, the supremum is the least upper bound for the set. The supremum does not need to be in the set. For $(0,1)$, the supremum is 1. This means that $1$ is an upper bound for $(0,1)$ [which is obvious] and that no number smaller than $1$ is an upper bound for $(0,1)$ [which follows from the above reasoning]. Whenever the maximum exists, it is equal to the supremum. Conversely, if the supremum lies in the set, then the maximum exists and is equal to this supremum.
The difference between minimum and infimum is similar.  In my example, $(0,1)$ has no minimum, but the infimum is 0. 
For your set, we see the sequence goes $$1, \,\,\,\,\, 2/3, \,\,\,\,\, 3/5, \,\,\,\,\, 4/7, \ldots.$$ The sequence is clearly decreasing. The number 1 is in the set and no other member of the set is larger than 1. Hence 1 is a maximum for the set. Since the maximum exists, it is equal to the supremum, so the supremum is also 1. There is no minimum element since for any element $x_n$ of the set, the value $x_{n+1}$ is less than $x_n$. However, the infimum is 1/2. To see this, we note that all members are greater than 1/2, so 1/2 is a lower bound. To be the infimum, it needs to be the greatest lower bound, so you need to prove that $1/2 + a$ is not a lower bound for any $a > 0$. Indeed, if $a > 0$, taking $n$ large enough that $1/(4n-2) < a$ shows that $$x_n = \frac{n}{2n -1} = \frac{n-1/2 + 1/2}{2n-1} = \frac 1 2 + \frac 1{4n-2} < \frac 1 2 + a.$$ Thus we have found a member of the sequence which is smaller than $1/2 + a$ so $1/2 + a$ is not a lower bound for the sequence. Hence $1/2$ is the infimum. 
A: Apparently, you know how to obtain the limits (here: upper bound 1, lower bound 1/2). That is any case is then the supremum (or infimum at the lower end) - or the set is unlimited, e.g. the naturals on top, the reals both ways: Nothing exists then.
Two cases:
(1) The limit is IN the set, e.g. actually attained by the sequence.
Then it is maximum AND supremum (or minimum AND infimum), e.g. 1.
(2) The limit is OUTSIDE the set, not attained by the sequence (though only barely missed). That is your case at the lower end, 1/2.
Then it is only supremum (or infimum) AND there is no maximum (minimum).
You might have some doubts coming in via the index n (your 3rd suggestion $n\to\infty$ and supremum)?
$n\to\infty$  has nothing to do with supremum ("large $n$"?). Only the $x_n$ themselves do.
Take-home bits:

*

*No bound - no ..mum, nothing.

*Minimum = Infimum OR no Minimum. (ditto max=sup or no max). In particular, these can not differ!

A: If the max exists, then it is the supremum. If the min exists, then it is the infimum. For an infinite set, it can happen that the max does not exist but the supremum does exist, and/or that the min does not exist but the infimum does exist. 
In this case, $\frac{n}{2n-1}$ is never actually equal to $\frac{1}{2}$, but $\frac{1}{2}$ is still the infimum, because it is a lower bound and because one can find $n$ with $\frac{n}{2n-1}<1/2+\epsilon$ whenever $\epsilon>0$ (which follows from the limit observation you made). Since the infimum is not attained, the minimum doesn't exist.
A: As Ian pointed out, the difference can only appear with infinite sets. Here is the definition of minimum and infimum:


*

*For the set $A$ to have a minimum $x_m$ it means that $\exists x_m\in A$ such that $x\geq x_m$ $\forall x\in A$.

*For the set $A$ to have an infimum $x_i$ it means that $\exists x_m$ such that $x\geq x_m$ $\forall x\in A$, and $x_m$ is the largest possible such number.


Notice that for the minimum, we require $x_m$ to still be in the set, while the infimum does not need to be in the set. That is the key point. For example, for $A=(0,\infty)$, we have that $x\geq 0$ $\forall x\in A$, and zero is the largest number that makes this inequality true for all elements of $A$. Since $0\notin A$, we cannot say that it is a minimum, so it can only be the infimum.
