Supremum and Infimum of this set.. 
Am I correct in taking the first few values for $m$ and $n$ as $1,2$ then $2,3$ then $3,4$ respectively? 
How do I go about finding the supremum and infimum for this sequence? 
Also how do I define the limit for this sequence and if I do will that help me find the supremum and infimum? 
I've been pulling my hair about this for a few hours now so any help on this topic would be greatly appreciated! Thanks!
 A: $m$ and $n$ can both be any natural number, so your job is to find the greatest lower bound, or inf,  (resp. least upper bound, sup) of the set for ANY values of $m$ and $n$. 
We can get an idea of how to do this by intuitively looking at what values of $m$ and $n$ should "maximize" and "minimize", respectively, the value of $2 + \frac{3}{m} - \frac{1}{n}$. 
To maximize $2 + \frac{3}{m} - \frac{1}{n}$, we want $\frac{3}{m}$ to be as big as possible and $\frac{1}{n}$ to be as small as possible. Since $m, n \in \mathbb{N}$, we quickly see that $m = 1$ maximizes the $\frac{3}{m}$ as $3$. But what about $-\frac{1}{n}$? Well, as $n$ gets larger and larger, $-\frac{1}{n}$ gets smaller and smaller. To get the supremum of this set, we want to look at the limit as $n$ goes to infinity, where $\frac{1}{n}$ approaches $0$. We therefore can make the smart guess that $5$ is the supremum of our set.  To formalize this, we must show that $5 - \frac{1}{n} < 5$ for all $n \in \mathbb{N}$, but we can always get within $\epsilon > 0$ of $5$ by choosing $n$ appropriately. The first is obvious. The second step I leave to you. 
Using the above as a guide, can you prove that the infimum of the set is $1$? 
A: First, $(m,n)$ could as well be $(17,42)$ or $(666,1)$ or $(25,25)$. The $m$ and $n$ are unrelated.
Try to make the expression $2+\frac3m-\frac1n$ big. To do so, make $\frac3m$ big and make $-\frac1n$ big. You may notice that $0<\frac3m\le3$ and $-1\le-\frac1n<0$. Hence $1=2+0-1<2+\frac3m-\frac1n<2+3+0=5$. Since you can get arbitrarily close to $1$ and $5$ respectively (take $n=1$ and $m$ very big or $m=1$ and $n$ very big), these numbers are the infimum and supremum.
A: Note that as $m$ increases, the term $a_{m,n} = 2+\frac3{m}-\frac1{n}$ decreases, and as $n$ increases the term $a_{m,n}$ decreases. So $m$ should be least possible - i.e. $m = 1$. At the same time, as $n$ tends to infinity, $a_{m,n}$ grows, but it will not reach its maximum. So the supremum of $\{a_{m,n}\}$ can be found by substituting $m = 1$ and taking the limit as $n$ tends to infinity. Similarly, the infimum is reached when $n = 1$ and the limit as $m$ tends to infinity is taken...
A: Write the sequence by $A_{m,n}$. First, $\inf_{m,n}A_{m,n}=1$. Since 
$$2+\frac{3}{m}-\frac{1}{n}\geq 2-\frac{1}{n}\geq 1$$
for any positive integers $m$ and $n$, $1$ is certainly a lower bound of $A_{m,n}$. Indeed, $1$ is the greatest lower bound of $A_{m,n}$. To see this we pick an arbitrary positive real number $\epsilon$. We may take $m=m_0$ large enough so that $3/m_0<\epsilon$, which implies that with $m=m_0$ and $n=1$ we have $$1<A_{m_0,1}=1+\frac{3}{m_0}<1+\epsilon.$$
On the other hand, $\sup_{m,n}A_{m,n}=5$. $5$ is an upper bound because
$$2+\frac{3}{m}-\frac{1}{n}\leq 5-\frac{1}{n}\leq 5.$$
However, it is the least upper bound of $A_{m,n}$. Take a sufficiently large $n=n_0$ so that $1/n_0<\epsilon$. Then, 
$$5-\epsilon<A_{1,n_0}=5-\frac{1}{n_0}<5$$
shows this fact.  
