Let us apply the results of § 69 to the particularly important case in which $$\phi(n) = x^{n}$$. If $$x = 1$$ then $$\phi(n) = 1$$, $$\lim\phi(n) = 1$$, and if $$x = 0$$ then $$\phi(n) = 0$$, $$\lim \phi(n) = 0$$, so that these special cases need not detain us.

First, suppose $$x$$ positive. Then, since $$\phi(n + 1) = x\phi(n)$$, $$\phi(n)$$ increases with $$n$$ if $$x > 1$$, decreases as $$n$$ increases if $$x < 1$$.

If $$x > 1$$, then $$x^{n}$$ must tend either to a limit (which must obviously be greater than $$1$$) or to $$+\infty$$. Suppose it tends to a limit $$l$$. Then $$\lim\phi(n + 1) = \lim\phi(n) = l$$, by Exs. XXV. 7; but $\lim\phi(n + 1) = \lim x\phi(n) = x\lim\phi(n) = xl,$ and therefore $$l = xl$$: and as $$x$$ and $$l$$ are both greater than $$1$$, this is impossible. Hence $x^{n} \to +\infty\quad (x > 1).$

Example. The reader may give an alternative proof, showing by the binomial theorem that $$x^{n} > 1 + n\delta$$ if $$\delta$$ is positive and $$x = 1 + \delta$$, and so that $x^{n} \to +\infty.$

On the other hand $$x^{n}$$ is a decreasing function if $$x < 1$$, and must therefore tend to a limit or to $$-\infty$$. Since $$x^{n}$$ is positive the second alternative may be ignored. Thus $$\lim x^{n} = l$$, say, and as above $$l = xl$$, so that $$l$$ must be zero. Hence $\lim x^{n} = 0\quad (0 < x < 1).$

Example. Prove as in the preceding example that $$(1/x)^{n}$$ tends to $$+\infty$$ if $$0 < x < 1$$, and deduce that $$x^{n}$$ tends to $$0$$.

We have finally to consider the case in which $$x$$ is negative. If $$-1 < x < 0$$ and $$x = -y$$, so that $$0 < y < 1$$, then it follows from what precedes that $$\lim y^{n} = 0$$ and therefore $$\lim x^{n} = 0$$. If $$x = -1$$ it is obvious that $$x^{n}$$ oscillates, taking the values $$-1$$, $$1$$ alternatively. Finally if $$x < -1$$, and $$x = -y$$, so that $$y > 1$$, then $$y^{n}$$ tends to $$+\infty$$, and therefore $$x^{n}$$ takes values, both positive and negative, numerically greater than any assigned number. Hence $$x^{n}$$ oscillates infinitely. To sum up: \begin{aligned} &\phi(n) = x^{n} \to +\infty &&(x > 1),\\ &\lim \phi(n) = 1 &&(x = 1),\\ &\lim \phi(n) = 0 &&(-1 < x < 1),\\ &\text{\phi(n) oscillates finitely} &&(x = -1),\\ &\text{\phi(n) oscillates infinitely}\qquad &&(x < -1).\end{aligned}

Example XXVII

1. If $$\phi(n)$$ is positive and $$\phi(n + 1) > K \phi(n)$$, where $$K > 1$$, for all values of $$n$$, then $$\phi(n) \to +\infty$$. These examples are particularly important and several of them will be made use of later in the text. They should therefore be studied very carefully.

[For $\phi(n) > K\phi(n – 1) > K^{2}\phi(n – 2) \dots > K^{n-1}\phi(1),$ from which the conclusion follows at once, as $$K^{n} \to\infty$$.]

2. The same result is true if the conditions above stated are satisfied only when $$n \geq n_{0}$$.

3. If $$\phi(n)$$ is positive and $$\phi(n + 1) < K\phi(n)$$, where $$0 < K < 1$$, then $$\lim\phi(n) = 0$$. This result also is true if the conditions are satisfied only when $$n \geq n_{0}$$.

4. If $$|\phi(n + 1)| < K|\phi(n)|$$ when $$n \geq n_{0}$$, and $$0 < K < 1$$, then $$\lim\phi(n) = 0$$.

5. If $$\phi(n)$$ is positive and $$\lim\{\phi(n + 1)\}/\{\phi(n)\} = l > 1$$, then $$\phi(n) \to +\infty$$.

[For we can determine $$n_{0}$$ so that $$\{\phi(n + 1)\}/\{\phi(n)\} > K > 1$$ when $$n \geq n_{0}$$: we may, e.g., take $$K$$ between $$1$$ and $$l$$. Now apply Ex. 1.]

6. If $$\lim\{\phi(n + 1)\}/\{\phi(n)\} = l$$, where $$l$$ is numerically less than unity, then $$\lim\phi(n) = 0$$. [This follows from Ex. 4 as Ex. 5 follows from Ex. 1.]

7. Determine the behaviour, as $$n \to \infty$$, of $$\phi(n) = n^{r}x^{n}$$, where $$r$$ is any positive integer.

[If $$x = 0$$ then $$\phi(n) = 0$$ for all values of $$n$$, and $$\phi(n) \to 0$$. In all other cases $\frac{\phi(n + 1)}{\phi(n)} = \left(\frac{n + 1}{n}\right)^{r}x \to x.$ First suppose $$x$$ positive. Then $$\phi(n) \to +\infty$$ if $$x > 1$$ (Ex. 5) and $$\phi(n) \to 0$$ if $$x < 1$$ (Ex. 6). If $$x = 1$$, then $$\phi(n) = n^{r} \to +\infty$$. Next suppose $$x$$ negative. Then $$|\phi(n)| = n^{r}|x|^{n}$$ tends to $$+\infty$$ if $$|x| \geq 1$$ and to $$0$$ if $$|x| < 1$$. Hence $$\phi(n)$$ oscillates infinitely if $$x \leq -1$$ and $$\phi(n) \to 0$$ if $$-1 < x < 0$$.]

8. Discuss $$n^{-r}x^{n}$$ in the same way. [The results are the same, except that $$\phi(n) \to 0$$ when $$x = 1$$ or $$-1$$.]

9. Draw up a table to show how $$n^{k}x^{n}$$ behaves as $$n \to \infty$$, for all real values of $$x$$, and all positive and negative integral values of $$k$$.

[The reader will observe that the value of $$k$$ is immaterial except in the special cases when $$x = 1$$ or $$-1$$. Since $$\lim\{(n + 1)/n\}^{k} = 1$$, whether $$k$$ be positive or negative, the limit of the ratio $$\phi(n + 1)/\phi(n)$$ depends only on $$x$$, and the behaviour of $$\phi(n)$$ is in general dominated by the factor $$x^{n}$$. The factor $$n^{k}$$ only asserts itself when $$x$$ is numerically equal to $$1$$.]

10. Prove that if $$x$$ is positive then $$\sqrt[n]{x} \to 1$$ as $$n \to \infty$$. [Suppose, , $$x > 1$$. Then $$x$$, $$\sqrt{x}$$, $$\sqrt[3]{x}$$, … is a decreasing sequence, and $$\sqrt[n]{x} > 1$$ for all values of $$n$$. Thus $$\sqrt[n]{x} \to l$$, where $$l \geq 1$$. But if $$l > 1$$ we can find values of $$n$$, as large as we please, for which $$\sqrt[n]{x} > l$$ or $$x > l^{n}$$; and, since $$l^{n} \to +\infty$$ as $$n \to \infty$$, this is impossible.]

11. $$\sqrt[n]{n}\to 1$$. [For $$\sqrt[n+1]{n + 1} < \sqrt[n]{n}$$ if $$(n + 1)^{n} < n^{n+1}$$ or $$\{1 + (1/n)\}^{n} < n$$, which is certainly satisfied if $$n \geq 3$$ (see § 73 for a proof). Thus $$\sqrt[n]{n}$$ decreases as $$n$$ increases from $$3$$ onwards, and, as it is always greater than unity, it tends to a limit which is greater than or equal to unity. But if $$\sqrt[n]{n}\to l$$, where $$l > 1$$, then $$n > l^{n}$$, which is certainly untrue for sufficiently large values of $$n$$, since $$l^{n}/n \to +\infty$$ with $$n$$ (Exs. 7, 8).]

12. $$\sqrt[n]{n!} \to +\infty$$. [However large $$\Delta$$ may be, $$n! > \Delta^{n}$$ if $$n$$ is large enough. For if $$u_{n} = \Delta^{n}/n!$$ then $$u_{n+1}/u_{n} = \Delta/(n + 1)$$, which tends to zero as $$n \to \infty$$, so that $$u_{n}$$ does the same (Ex. 6).]

13. Show that if $$-1 < x < 1$$ then $u_{n} = \frac{m(m – 1) \dots (m – n + 1)}{n!} x^{n} = \binom{m}{n} x^{n}$ tends to zero as $$n \to \infty$$.

[If $$m$$ is a positive integer, $$u_{n} = 0$$ for $$n > m$$. Otherwise $\frac{u_{n+1}}{u_{n}} = \frac{m – n}{n + 1}x \to -x,$ unless $$x = 0$$.]