Consider the behaviour as \(n\) tends to \(\infty\) of the following functions:
1. \((-1)^{n}\), \(5 + 3(-1)^{n}\), \((1,000,000/n) + (-1)^{n}\), \(1,000,000(-1)^{n} + (1/n)\).
2. \((-1)^{n}n\), \(1,000,000 + (-1)^{n}n\).
3. \(1,000,000 – n\), \((-1)^{n}(1,000,000 – n)\).
4. \(n\{1 + (-1)^{n}\}\). In this case the values of \(\phi(n)\) are \[0,\quad 4,\quad 0,\quad 8,\quad 0,\quad 12,\quad 0,\quad 16,\ \dots.\] The odd terms are all zero and the even terms tend to \(+\infty\): \(\phi(n)\) oscillates infinitely.
5. \(n^{2} + (-1)^{n}2n\). The second term oscillates infinitely, but the first is very much larger than the second when \(n\) is large. In fact \(\phi(n) \geq n^{2} – 2n\) and \(n^{2} – 2n = (n – 1)^{2} – 1\) is greater than any assigned value \(\Delta\) if \(n > 1 + \sqrt{\Delta + 1}\). Thus \(\phi(n) \to +\infty\). It should be observed that in this case \(\phi(2k + 1)\) is always less than \(\phi(2k)\), so that the function progresses to infinity by a continual series of steps forwards and backwards. It does not however ‘oscillate’ according to our definition of the term.
6. \(n^{2}\{1 + (-1)^{n}\}\), \((-1)^{n}n^{2} + n\), \(n^{3} + (-1)^{n}n^{2}\).
7. \(\sin n\theta\pi\). We have already seen (Exs. XXIII. 9) that \(\phi(n)\) oscillates finitely when \(\theta\) is rational, unless \(\theta\) is an integer, when \(\phi(n)= 0\), \(\phi(n) \to 0\).
The case in which \(\theta\) is irrational is a little more difficult. But it is not difficult to see that \(\phi(n)\) still oscillates finitely. We can without loss of generality suppose \(0 < \theta < 1\). In the first place \(|\phi(n)| < 1\). Hence \(\phi(n)\) must oscillate finitely or tend to a limit. We shall consider whether the second alternative is really possible. Let us suppose that \[\lim \sin n\theta\pi = l.\] Then, however small \(\epsilon\) may be, we can choose \(n_{0}\) so that \(\sin n\theta\pi\) lies between \(l -\epsilon\) and \(l + \epsilon\) for all values of \(n\) greater than or equal to \(n_{0}\). Hence \(\sin(n + 1)\theta\pi – \sin n\theta\pi\) is numerically less than \(2\epsilon\) for all such values of \(n\), and so \(|\sin \frac{1}{2}\theta\pi \cos(n + \frac{1}{2})\theta\pi| < \epsilon\).
Hence \[\cos(n + \tfrac{1}{2})\theta\pi = \cos n\theta\pi \cos\tfrac{1}{2}\theta\pi – \sin n\theta\pi \sin\tfrac{1}{2}\theta\pi\] must be numerically less than \(\epsilon/|\sin\frac{1}{2}\theta\pi|\). Similarly \[\cos(n – \tfrac{1}{2})\theta\pi = \cos n\theta\pi \cos\tfrac{1}{2}\theta\pi + \sin n\theta\pi \sin\tfrac{1}{2}\theta\pi\] must be numerically less than \(\epsilon/|\sin\frac{1}{2}\theta\pi|\); and so each of \(\cos n\theta\pi \cos\frac{1}{2}\theta\pi\), \(\sin n\theta\pi \sin\frac{1}{2}\theta\pi\) must be numerically less than \(\epsilon/|\sin\frac{1}{2}\theta\pi|\). That is to say, \(\cos n\theta\pi \cos\frac{1}{2}\theta\pi\) is very small if \(n\) is large, and this can only be the case if \(\cos n\theta\pi\) is very small. Similarly \(\sin n\theta\pi\) must be very small, so that \(l\) must be zero. But it is impossible that \(\cos n\theta\pi\) and \(\sin n\theta\pi\) can both be very small, as the sum of their squares is unity. Thus the hypothesis that \(\sin n\theta\pi\) tends to a limit \(l\) is impossible, and therefore \(\sin n\theta\pi\) oscillates as \(n\) tends to \(\infty\).
The reader should consider with particular care the argument ‘\(\cos n\theta\pi \cos\frac{1}{2}\theta\pi\) is very small, and this can only be the case if \(\cos n\theta\pi\) is very small’. Why, he may ask, should it not be the other factor \(\cos\frac{1}{2}\theta\pi\) which is ‘very small’? The answer is to be found, of course, in the meaning of the phrase ‘very small’ as used in this connection. When we say ‘\(\phi(n)\) is very small’ for large values of \(n\), we mean that we can choose \(n_{0}\) so that \(\phi(n)\) is numerically smaller than any assigned number, if . Such an assertion is palpably absurd when made of a fixed number such as \(\cos\frac{1}{2}\theta\pi\), which is not zero.
Prove similarly that \(\cos n\theta\pi\) oscillates finitely, unless \(\theta\) is an even integer.
8. \(\sin n\theta\pi + (1/n)\), \(\sin n\theta\pi + 1\), \(\sin n\theta\pi + n\), \((-1)^{n} \sin n\theta\pi\).
9. \(a\cos n\theta\pi + b\sin n\theta\pi\), \(\sin^{2}n\theta\pi\), \(a\cos^{2}n\theta\pi + b\sin^{2}n\theta\pi\).
10. \(a + bn + (-1)^{n} (c + dn) + e\cos n\theta\pi + f\sin n\theta\pi\).
11. \(n\sin n\theta\pi\). If \({\theta}\) is integral, then \(\phi(n) = 0\), \(\phi(n) \to 0\). If \(\theta\) is rational but not integral, or irrational, then \(\phi(n)\) oscillates infinitely.
12. \(n(a\cos^{2} n\theta\pi + b\sin^{2} n\theta\pi)\). In this case \(\phi(n)\) tends to \(+\infty\) if \(a\) and \(b\) are both positive, but to \(-\infty\) if both are negative. Consider the special cases in which \(a = 0\), \(b > 0\), or \(a > 0\), \(b = 0\), or \(a = 0\), \(b = 0\). If \(a\) and \(b\) have opposite signs \(\phi(n)\) generally oscillates infinitely. Consider any exceptional cases.
13. \(\sin(n^{2}\theta\pi)\). If \(\theta\) is integral, then \(\phi(n) \to 0\). Otherwise \(\phi(n)\) oscillates finitely, as may be shown by arguments similar to though more complex than those used in Exs. XXIII. 9 and Exs. XXIV. 7.
14. \(\sin(n!\, \theta\pi)\). If \(\theta\) has a rational value \(p/q\), then \(n!\, \theta\) is certainly integral for all values of \(n\) greater than or equal to \(q\). Hence \(\phi(n) \to 0\). The case in which \(\theta\) is irrational cannot be dealt with without the aid of considerations of a much more difficult character.
15. \(\cos(n!\, \theta\pi)\), \(a\cos^{2}(n!\, \theta\pi) + b\sin^{2}(n!\, \theta\pi)\), where \(\theta\) is rational.
16. \(an – [bn]\), \((-1)^{n}(an – [bn])\).
17. \([\sqrt{n}]\), \((-1)^{n}[\sqrt{n}]\), \(\sqrt{n} – [\sqrt{n}]\).
18. The smallest prime factor of \(n\). When \(n\) is a prime, \(\phi(n) = n\). When \(n\) is even, \(\phi(n) = 2\). Thus \(\phi(n)\) oscillates infinitely.
19. The largest prime factor of \(n\).
20. The number of days in the year \(n\) A.D.
