infimum and supremum of n/(n+k) where n,k are natural numbers

Question

find infimum and supremum of set $\frac n{n+k}$ where n,k are natural numbers

firstly i wrote $$\frac{n}{n+k} = \frac{n+k-k}{n+k}$$ which is equal to $$1-\frac k{n+k}$$ and we know that $n$ and $k$ are natural numbers so supremum is 1

i know that infimum of this set is 0 but i dont know how can i show this

i tried to show this by some epsilon not just writing that n is always smaller than n+k and the gap between them increases but its always greater than 0 for my class this is not enough

i would be very thankful if someone could explain this to me

Answer 1

$\frac n {n+k}$ is always $>0$ so infimum is $\ge 0$.

Then $\forall \varepsilon > 0, \exists n, k \in \mathbb{N}, \frac n {n+k} < \varepsilon \;$:
choose $n=1$ and $k=\lceil \frac 1 {\varepsilon} \rceil$ for example.

Answer 2

Your guess that the infimum is $0$ is correct. “Infimum” means “greatest lower bound”. To prove that $0$ is the infimum, you need to show:

1. $0$ is a lower bound. That is, for any $n$ and $k$, $$0≤ \frac n{n+k}.$$

2. $0$ is the greatest lower bound: for any $\epsilon > 0$, then $\epsilon$ is not a lower bound for the set. That is, there is $n$ and $k$ so that $$\frac n{n+k}<\epsilon.$$

Does that help?