Quoting: find the supremum and infimum of S. State whether there are in S
$$ S=\{x|x=-(1/n)+[1+(-1)^n]n^2,n \geq 1\}$$
I am new to new to this.
What I did is substituting some values for n:
When $n = 1$ , then $ x= -1$
When $n = 2$ . then $x = -1/2 +8$
When $n = 3$ , then $x = -1/3$
When $n = 4$ , then $x = -1/4 +32$
Looking at the pattern, It seems to me that we have as lower bounds: any number less than -1, infimum -1
there is no upper bound, $Sup(S)= \infty$ .
Is this correct? Is there another way a bit more efficient?
Much appreciated
Best Answer
You're on the right track, but plugging in values isn't enough for a rigorous proof. When $n = 1$, $x = -1$, so $\inf(S)\leq -1$. Also, $$-\frac{1}{n}+(1+(-1)^n)n^2\geq -\frac{1}{n}\geq -1$$ so $-1$ is a lower bound for $S$. Therefore, $\inf(S) = -1$. Next, letting $n$ be even, we have $n = 2k$ for some $k\geq 1$. Then, $$-\frac{1}{n}+(1+(-1)^n)n^2 = -\frac{1}{2k}+8k^2$$ and $$\lim_{k\to \infty} \left[-\frac{1}{2k}+8k^2\right] = \infty$$ which implies that for any $y$, there is some $k$ such that $-\frac{1}{2k}+8k^2 > y$. Thus, there is no finite upper bound for $S$, so $\sup(S) = \infty$. Therefore, $S$ contains its infimum, but not its supremum.