What is the supremum and infimum of $n/(1+n^2)$ where $n$ is an element of $\mathbb{N}$?

Question

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?

Answer 1

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 $\displaystyle y=\frac n{1+n^2}$

Answer 2

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 $n$

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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 $\displaystyle \frac n{n^2+1}$?