What is the supremum and infimum of $n/(1+n^2)$ where $n$ is an element of $\mathbb{N}$? Please help with the proving! I would like to double check my answers.
$\sup = 1/2$
$\max = 1/2$
$\inf = 0$
$\min = \text{No minimum}$
Is this correct? How do I go about showing that no smaller number is also an upper bound?
 A: As $n$ is real, we have $\displaystyle(n-1)^2\ge0\implies n^2+1\ge2n\implies \frac12\ge \frac n{n^2+1}$ the equality occurs if $n=1$
Observe that the value of $\displaystyle \frac n{n^2+1}$ goes arbitrarily close to zero as $n\to\infty$
So, there will be no minimum value of $\displaystyle \frac n{n^2+1}$ for finite $n$

Alternatively, let us consider the reciprocal of $\displaystyle \frac n{n^2+1}$ i.e., $\displaystyle \frac{n^2+1}n=n+\frac1n$
As $n>0$ using $A.M.\ge G.M.n+\frac1n\ge 2\sqrt{n\cdot\frac1n}=2$ and it has evidently no maximum value
If the reciprocal has minimum value $(=2)$ only,  what should be the fact of the original function $\displaystyle \frac n{n^2+1}$?
A: Put $\displaystyle\frac n{1+n^2}=y\iff yn^2-n+y=0$
As $n$ is real, the discriminant of the above quadratic equation in $n$  must be $\ge0$
this will give us the range of $\displaystyle y=\frac n{1+n^2}$
A: Your problem implicitly leads to the question whether zero is or not an element of N. About this question there is not general agreement among the mathematicians of the world. If yes, then there is a minimun (which is contrary to your result)
