It will be convenient to prove at this stage a number of elementary inequalities which will be useful to us later on.

(i) It is evident that if $$\alpha > 1$$ and $$r$$ is a positive integer then $r\alpha^{r} > \alpha^{r-1} + \alpha^{r-2} + \dots + 1.$ Multiplying both sides of this inequality by $$\alpha – 1$$, we obtain $r\alpha^{r}(\alpha – 1) > \alpha^{r} – 1;$ and adding $$r(\alpha^{r} – 1)$$ to each side, and dividing by $$r(r + 1)$$, we obtain $\begin{equation*} \frac{\alpha^{r+1} – 1}{r + 1} > \frac{\alpha^{r} – 1}{r}\quad (\alpha > 1). \tag{1} \end{equation*}$ Similarly we can prove that $\begin{equation*} \frac{1 – \beta^{r+1}}{r + 1} < \frac{1 – \beta^{r}}{r}\quad (0 < \beta < 1). \tag{2} \end{equation*}$

It follows that if $$r$$ and $$s$$ are positive integers, and $$r > s$$, then $\begin{equation*} \frac{\alpha^{r} – 1}{r} > \frac{a^{s} – 1}{s},\quad \frac{1 – \beta^{r}}{r} < \frac{1 – \beta^{s}}{s}. \tag{3} \end{equation*}$ Here $$0 < \beta < 1 < \alpha$$. In particular, when $$s = 1$$, we have $\begin{equation*} \alpha^{r} – 1 > r(\alpha – 1),\quad 1 – \beta^{r} < r(1 – \beta). \tag{4} \end{equation*}$

(ii) The inequalities (3) and (4) have been proved on the supposition that $$r$$ and $$s$$ are positive integers. But it is easy to see that they hold under the more general hypothesis that $$r$$ and $$s$$ are any positive rational numbers. Let us consider, for example, the first of the inequalities (3). Let $$r = a/b$$, $$s = c/d$$, where $$a$$, $$b$$, $$c$$, $$d$$ are positive integers; so that $$ad > bc$$. If we put $$\alpha = \gamma^{bd}$$, the inequality takes the form $(\gamma^{ad} – 1)/ad > (\gamma^{bc} – 1)/bc;$ and this we have proved already. The same argument applies to the remaining inequalities; and it can evidently be proved in a similar manner that $\begin{equation*} \alpha^{s} – 1 < s(\alpha – 1),\quad 1 – \beta^{s} > s(1 – \beta), \tag{5} \end{equation*}$ if $$s$$ is a positive rational number less than $$1$$.

(iii) In what follows it is to be understood that all the letters denote positive numbers, that $$r$$ and $$s$$ are rational, and that $$\alpha$$ and $$r$$ are greater than $$1$$, $$\beta$$ and $$s$$ less than $$1$$. Writing $$1/\beta$$ for $$\alpha$$, and $$1/\alpha$$ for $$\beta$$, in (4), we obtain $\begin{equation*} \alpha^{r} – 1 < r\alpha^{r-1}(\alpha – 1),\quad 1 – \beta^{r} > r\beta^{r-1}(1 – \beta). \tag{6} \end{equation*}$ Similarly, from (5), we deduce $\begin{equation*} \alpha^{s} – 1 > s\alpha^{s-1}(\alpha – 1),\quad 1 – \beta^{s} < s\beta^{s-1}(1 – \beta). \tag{7} \end{equation*}$

Combining (4) and (6), we see that $\begin{equation*} r\alpha^{r-1}(\alpha – 1) > \alpha^{r} – 1 > r(\alpha – 1). \tag{8} \end{equation*}$ Writing $$x/y$$ for $$\alpha$$, we obtain $\begin{equation*} rx^{r-1} (x – y) > x^{r} – y^{r} > ry^{r-1} (x – y) \tag{9} \end{equation*}$ if $$x > y > 0$$. And the same argument, applied to (5) and (7), leads to $\begin{equation*} sx^{s-1} (x – y) < x^{s} – y^{s} < sy^{s-1} (x – y). \tag{10} \end{equation*}$

Example XXVIII

1. Verify (9) for $$r = 2$$, $$3$$, and (10) for $$s = \frac{1}{2}$$, $$\frac{1}{3}$$.

2. Show that (9) and (10) are also true if $$y > x > 0$$.

3. Show that (9) also holds for $$r < 0$$. [See Chrystal’s Algebra, vol. ii, pp. 43–45.]

4. If $$\phi(n) \to l$$, where $$l > 0$$, as $$n \to \infty$$, then $$\phi^{k} \to l^{k}$$, $$k$$ being any rational number.

[We may suppose that $$k > 0$$, in virtue of Theorem III of § 66; and that $$\frac{1}{2}l < \phi < 2l$$, as is certainly the case from a certain value of $$n$$ onwards. If $$k > 1$$, $k\phi^{k-1}(\phi – l) > \phi^{k} – l^{k} > kl^{k-1}(\phi – l)$ or $kl^{k-1}(l – \phi) > l^{k} – \phi^{k} > k\phi^{k-1}(l – \phi),$ according as $$\phi > l$$ or $$\phi < l$$. It follows that the ratio of $$|\phi^{k} – l^{k}|$$ and $$|\phi – l|$$ lies between $$k(\frac{1}{2}l)^{k-1}$$ and $$k(2l)^{k-1}$$. The proof is similar when $$0 < k < 1$$. The result is still true when $$l = 0$$, if $$k > 0$$.]

5. Extend the results of Exs. XXVII. 7, 8, 9 to the case in which $$r$$ or $$k$$ are any rational numbers.