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Let $S = \{\frac{(-1)^n}{n} : n \in \mathbb{N}\}$ and $f(x) = \exp(x)$. Prove that $f(S)$ is bounded from above and below and find its $\sup f(S)$ and $\inf f(S)$. Are these values equal to the maximum and minimum of $f$ on $S$?

My attempt: So $f(S)$ is the set $\{f(x) : x \in S\} = \{\exp(x) : x \in S\} = \{\exp(\frac{(-1)^n}{n}) : n \in \mathbb{N}\}$. To prove that $f(S)$ is bounded above, I need to show that there exists a $u \in \mathbb{R}$ such that $y \le u$ for all $y \in f(S)$. And to prove that $f(S)$ is bounded below, I need to show that there exists a $l \in \mathbb{R}$ such that $l \le y$ for all $y \in f(S)$. How can I find such $u$ and $l$? Furthermore, how can I then find the supremum and infimum and prove those values are indeed the supremum and infimum? Likewise for the maximum and minimum.

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    $\begingroup$ You can find such $u$ and $l$ by asking yourself: how large and how small can $\exp((-1)^n/n)$ be when $n$ is a naturaly number? You may want to use the fact that $e^x$ is monotonically increasing, and that $(-1)^n/n<1$ for all $n$, and that $(-1)^n/n>-1$ for all $n$, for example. $\endgroup$ – uniquesolution Jul 20 '17 at 9:38
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Well, note that $f$ is strictly increasing over $\mathbb{R}$. Also note that $S$ has both a maximum and a minimum element. To see this, at first, let us write down a list of some of $S$'s elements: $$S=\left\{\frac{(-1)^n}{n}:n\in\mathbb{N}\right\}=\left\{-1,-\frac{1}{3},-\frac{1}{5},\dots,\frac{1}{6},\frac{1}{4},\frac{1}{2}\right\}=\left\{-1,-\frac{1}{3},\dots\right\}\cup\left\{\frac{1}{2},\frac{1}{4},\dots\right\}$$

So, now it is evident that $S$'s minimum is $-1$ and its maximum is $\frac{1}{2}$.

Now, it is obvious that: $$\max f(S)=\sup f(S)=f\left(\frac{1}{2}\right)=\sqrt{e}$$ and $$\min f(S)=\inf f(S)=f\left(-1\right)=\frac{1}{e}$$

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