Calculate $\sup \,\left\{\frac{n}{n+1}:n \in\mathbb{N}\right\}$

Question

Calculate $\sup \,\left\{\frac{n}{n+1}:n \in\mathbb{N}\right\}$

Can anyone help me with this? I am very confused with this question. Thank you.

Answer 1

Let $A\neq \phi$. Then Sup $A=a$ iff

1) $x\leq a$ for all $x\in A$ and

2) Given $\epsilon >0$, there exists $x_A \in A$ such that $x_A > a- \epsilon$.

So, here $\frac{n}{n+1}\leq 1$ for each $n$. Let $\epsilon >0$. Then there exists $N\in \mathbb{N}$ such that $N> \frac{1}{\epsilon}$.

$\Rightarrow N+1> \frac{1}{\epsilon}$

$\Rightarrow 1- \epsilon< \frac{N}{N+1}$. Hence by definition Sup is $1$.

Answer 2

Hint:Obviously $\frac{n}{n+1} \leq 1.$ Moreover for large values of $n$ we observe that $\frac{n}{n+1}$ is so close to $1$.(you can convince yourself using $\epsilon-$definition)in other words $\lim_{n\to\infty}\frac{n}{n+1}=1$.

Assume $\epsilon>0$ is given then there exists $n\in\mathbb N$ such that $\frac{n}{n+1}+\epsilon\geq 1.$ Proof by contradiction: Suppose $\frac{n}{n+1}+\epsilon<1$ for all natural numbers $n$, then $n+\epsilon(n+1)<n+1\Rightarrow \epsilon(n+1)<1\Rightarrow n+1<\frac{1}{\epsilon}\,\,\forall n\in\mathbb N!!!$ But this is a contradiction

Answer 3

Just write down some numbers of this set (sequence): if we set $S = \{ \frac{n}{n+1} : n \in \mathbb{N} \}$, we get \begin{equation} S = \{ (0, \text{ if you let } 0 \in \mathbb{N}), \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, ..., \frac{100000}{100001}, ... \} \end{equation}

You can see that the "biggest" number you will get is just $1$, the limit of the sequence $\frac{n}{n+1}$. Prove that indeed $1$ is the supremum of this set:

· Obviously $1$ is greater than any number in $S$.

· For every $\epsilon > 0$, you will always find a number $\frac{N}{N+1} \in S$ such that $1-\epsilon \leq \frac{N}{N+1} < 1$ because of the definition of a limit (as we said, $1$ is the limit of the sequence).

Hence $1 = sup(S)$.