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