Proof to calculate supremum

Question

I was wondering if someone could help me with a proof for the supremum of the set $E = \frac{1}{n} + (-1)^n$ where n is a natural number. I can see that the supremum is equal to 3/2 at n=1 but wasnt sure how to justify it. previously i have seen epsilon values but dont see how to do that here. maybe could use subsets and look at even terms? thanks

Answer 1

$\frac 1 n +(-1)^{n}\leq \frac 1 n +1 <\frac 3 2$ when $n >2$ and $\frac 1 n +(-1)^{n}=0$ when $n=1$. Hence the maximum value of $\frac 3 2$ is attained when $n=2$.

Answer 2

First we prove that $3/2$ is an upper bound for $E$.
$E(n=1)=0\lt 3/2$
$E(n=2)=1/2+(-1)^2=3/2\le 3/2$
For $n\ge 3, E(n)=1/n+(-1)^n\lt1/2+1=3/2$
Therefore, $E(n)\le 3/2$ for all $n\in \mathbb N \implies 3/2$ is an upper bound for $E$.

Suppose on the contrary that $3/2$ is not a supremum then there exists $\epsilon_0\gt 0$ such that $ E(n)\lt 3/2-\epsilon_0$ for all $n\in \mathbb N$
But clearly, $3/2-\epsilon_0\lt 3/2=1/2+(-1)^2=E(2)$, which is a contradiction and hence $3/2$ is a supremum of $E$ in set of real nos.