I'm trying to prove that if $b > 1$ then $\sup\{b^{-1/n} : n\in\mathbb{N}\} = 1$. It is part of the solution to exercise 6d in Principles of Mathematical Analysis on page 22 so I can't use any tools not introduced on pages prior. My work so far (lots of help from the internet):
If $b > 1$ then $b^{1/n} > 1$ and $b^{-1/n} < 1$. So 1 is an upper bound for $\{b^{-1/n} : n\in\mathbb{N}\}$.
Next task is to show that no number smaller than 1 is an upper bound for the set. We'll use proof by contradiction. Introduce $0 < \epsilon < 1$ and $\delta > 0$ related so that $1 - \epsilon = \frac{1}{1 + \delta}$. Raising both sides by $n$ $$ (1 - \epsilon)^n = \frac{1}{(1 + \delta)^n} \leq \frac{1}{1+n\delta} $$ for sufficiently large $n$. That makes it clear that $(1 - \epsilon)^n < b$ holds and therefore $1 - \epsilon < b^{1/n}$. But how to proceed? I want to show that $1 - \epsilon < b^{-1/n}$.
Edit: Fixed proof:
Proof by contradiction shows that there can be no upper bound smaller than 1 for $B$. Suppose that there is an $0 < \alpha < 1$ so that which is an upper bound for $B$. Then $$ \frac{1}{b^{1/n}} < \alpha \iff \frac{1}{\alpha^n} < b. $$ If we pick an arbitrary number so that $b = 35$, we get $$ \frac{1}{\alpha^n} < 35. $$ Then it is clear that for any $0 < \alpha < 1$, we can pick a large enough $n$ to falsify the inequality. Therefore we have a contradiction and $\sup B$ must be 1.
The sequence $(b^{-1/n})$ is increasing (take the logarithm). Since it's bounded, it converges to its supremum. But the sequence converges to 1 (because its logarithm converges to zero). Therefore the sup is 1.