72. The limit of $x^n$

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 $x = 0$.]

$\leftarrow$ 71. Alternative proof of Weierstrass’s Theorem

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73.The limit of $(1+\frac{1}{n})^n$ $\rightarrow$