Elementary proof that $a_n \to a \implies a_n^r \to a^r$ for $r \in \mathbb{Q}$ I've been looking for an elementary proof of the fact that rational exponentiation is continuous (for sequences). Let $r = p/q \in \mathbb{Q},$ with $p,q \in \mathbb{N}.$ Assuming $|a_n - a| \to 0,$ we have
$$|a_n^r - a^r| = |(a_n^{1/q})^p - (a^{1/q})^p| = \left|a_n^{1/q} - a^{1/q}\right|\left|\sum_{i=0}^{p-1} a_n^{p-1-i}a^i\right|.$$
Is there a way to bound this further in terms of $|a_n - a|$ to conclude?
Thank you.
 A: We need to establish the following inequalities $$rx^{r - 1}(x - y) > x^{r} - y^{r} > ry^{r - 1}(x - y)$$ and $$sx^{s - 1}(x - y) < x^{s} - y^{s} < sy^{s - 1}(x - y)$$ where $x > y > 0$ and $r > 1, 0 < s < 1$ with $r, s \in \mathbb{Q}$. Also these inequalities need to be established via algebra. I will establish here the inequalities related to $r > 1$.
Let $\alpha > 1$ be a real number and assume $r$ is a positive integer. Then we can easily see that $$r\alpha^{r} > \alpha^{r - 1} + \alpha^{r - 2} + \cdots + 1$$ so that $$r\alpha^{r}(\alpha - 1) > \alpha^{r} - 1$$ Adding $r(\alpha^{r} - 1)$ on each side we get $$r(\alpha^{r + 1} - 1) > (r + 1)(\alpha^{r} - 1)$$ and finally $$\frac{\alpha^{r + 1} - 1}{r + 1} > \frac{\alpha^{r} - 1}{r}$$ It thus follows that the expression $\dfrac{\alpha^{r} - 1}{r}$ increases with $r$ and therefore if $r, s$ are positive integers with $r > s$ then $$\frac{\alpha^{r} - 1}{r} > \frac{\alpha^{s} - 1}{s}\,\,\,\cdots (1)$$ It follows that if $r$ is a positive integer and $r > 1$ then $$\frac{\alpha^{r} - 1}{r} > \frac{\alpha^{1} - 1}{1}$$ so that $$\alpha^{r} - 1 > r(\alpha - 1)\,\,\,\cdots (2)$$ Similarly we can show that if $0 < \beta < 1$ then $$\frac{1 - \beta^{r}}{r} < \frac{1 - \beta^{s}}{s}\,\,\,\cdots (3)$$ and $$1 - \beta^{r} < r(1 - \beta)\,\,\,\cdots (4)$$ if $r > 1$.
The inequalities above have been proved with restriction that $r, s$ be positive integers. This can be easily extended to the case when $r, s$ are rational. Let $r = a/b, s = c/d$ where $a, b, c, d$ are positive integers and then $r > s$ implies that $ad > bc$. Let $\gamma = \alpha^{1/bd}$ so that $\alpha = \gamma^{bd}$ then $\gamma > 1$ and hence by $(1)$ we get $$\frac{\gamma^{ad} - 1}{ad} > \frac{\gamma^{bc} - 1}{bc}$$ or $$\frac{\alpha^{r} - 1}{r} > \frac{\alpha^{s} - 1}{s}$$ Similarly all the inequalities $(2), (3), (4)$ can be extended to rational values of $r, s$.
Next if $\alpha > 1$ then $0 < 1/\alpha < 1$ and $0 < \beta < 1$ implies $1/\beta > 1$. Therefore writing $1/\beta$ for $\alpha$ in $(2)$ and $1/\alpha$ for $\beta$ in $(4)$ we get $$\alpha^{r} - 1 < r\alpha^{r - 1}(\alpha - 1)\,\,\,\cdots (5)$$ and $$1 - \beta^{r} > r\beta^{r - 1}(1 - \beta)\,\,\,\cdots (6)$$ If $x > y > 0$ then we can put $\alpha = x/y$ and $\beta = y/x$ in the above inequalities and get $$rx^{r - 1}(x - y) > x^{r} - y^{r} > ry^{r - 1}(x - y)$$
Using these inequalities it is possible to bound $|a_{n}^{r} - a^{r}|$ in terms of $|a_{n} - a|$ and you can complete your proof easily.
Update: This is essentially the elementary technique used to prove continuity of $f(x) = x^{r}$ for $x > 0$ and $r \in \mathbb{Q}$ and if you notice carefully it can also be used to establish the differentiation formula $(x^{r})' = rx^{r - 1}$.
