Exercise 2.5.6

To show $b^{1∕n}$ is monotone bounded consider two cases (I won’t prove each rigorously to avoid clutter, you can if you want)

(i)

If $b>1$ then $b^{1∕n}$ is decreasing, and bounded $b^{1∕n}>1⟺b>1^n$ (raise both to $n$)

(ii)

If $b<1$ then $b^{1∕n}$ is increasing, and bounded $b^{1∕n}<1⟺b<1^n$ (raise both to $n$)

therefore $b^{1∕n}$ converges for each $b\ge 0$ by the monotone convergence theorem. To find the limit, $\lim <!--\; FUNCTION\; APPLICATION\; -->b^{1∕n}=c$, consider the subsequence:

From Exercise 2.3.1 we know that $\lim <!--\; FUNCTION\; APPLICATION\; -->\sqrt{b^{1∕n}}=\sqrt{c}$ and since subsequences of a convergent sequence converge to the same limit:

For the case $b>1$ (and therefore $b^{1∕n}>1$) the only possible limit is $c=1$. For $b<1$ the increasing sequence is bounded by $b^{1∕n}\ge b$, excluding the other solution ( $c=0$) for all values of b except for $b=0$, whose limit can be determined directly as $\lim <!--\; FUNCTION\; APPLICATION\; -->0^{1∕n}=\lim <!--\; FUNCTION\; APPLICATION\; -->0=0$.