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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

johnsdgh
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    The epsilon arguments are usually used for limit type supremums/infimum (e.g. $\inf E = -1$ in this case) but since in this case the supremum is actually attained, you only need to show that $\frac 1 n + (-1)^n \le \frac 3 2$ for all natural numbers $n$. – player3236 Sep 11 '20 at 06:42
  • The idea can be separation to $$E_{2n}\E_{2n+1}$$ – Khosrotash Sep 11 '20 at 06:43

2 Answers2

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

Kavi Rama Murthy
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  • Is this considered enough justification for it? This is for an intro analysis course and i was told to justify my answer and i wasnt sure if this was considered enough – johnsdgh Sep 11 '20 at 06:47
  • Yes, it is enough. When there is a maximal element in a set that element is the supremum and no further justification is required. @johnsdgh – Kavi Rama Murthy Sep 11 '20 at 07:23
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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.

Koro
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