Definition. When \(\phi(n)\) does not tend to a limit, nor to \(+\infty\), nor to \(-\infty\), as \(n\) tends to \(\infty\), we say that \(\phi(n)\) oscillates as \(n\) tends to \(\infty\).

A function \(\phi(n)\) certainly oscillates if its values form, as in the case considered in the last example above, a continual repetition of a cycle of values. But of course it may oscillate without possessing this peculiarity. Oscillation is defined in a purely negative manner: a function oscillates when it does not do certain other things.

The simplest example of an oscillatory function is given by \[\phi(n) = (-1)^{n},\] which is equal to \(+1\) when \(n\) is even and to \(-1\) when \(n\) is odd. In this case the values recur cyclically. But consider \[\phi(n) = (-1)^{n} + (1/n),\] the values of which are \[-1 + 1,\quad 1 + (1/2),\quad -1 + (1/3),\quad 1 + (1/4),\quad -1 + (1/5),\ \dots.\] When \(n\) is large every value is nearly equal to \(+1\) or \(-1\), and obviously \(\phi(n)\) does not tend to a limit or to \(+\infty\) or to \(-\infty\), and therefore it oscillates: but the values do not recur. It is to be observed that in this case every value of \(\phi(n)\) is numerically less than or equal to \(3/2\). Similarly \[\phi(n) = (-1)^{n} 100 + (1000/n)\] oscillates. When \(n\) is large, every value is nearly equal to \(100\) or to \(-100\). The numerically greatest value is \(900\) (for \(n = 1\)). But now consider \(\phi(n) = (-1)^{n}n\), the values of which are \(-1\), \(2\), \(-3\), \(4\), \(-5\), …. This function oscillates, for it does not tend to a limit, nor to \(+\infty\), nor to \(-\infty\). And in this case we cannot assign any limit beyond which the numerical value of the terms does not rise. The distinction between these two examples suggests a further definition.

Definition. If \(\phi(n)\) oscillates as \(n\) tends to \(\infty\), then \(\phi(n)\) will be said to oscillate finitely or infinitely according as it is or is not possible to assign a number \(K\) such that all the values of \(\phi(n)\) are numerically less than \(K\), \(|\phi(n)| < K\) for all values of \(n\).

These definitions, as well as those of § 58 and 60, are further illustrated in the following examples.

Example XXIV

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. 7.1

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.

 

Example XXV

1. If \(\phi(n) \to +\infty\) and \(\psi(n) \geq \phi(n)\) for all values of \(n\), then \(\psi(n) \to +\infty\).

2. If \(\phi(n) \to 0\), and \(|\psi(n)| \leq |\phi(n)|\) for all values of \(n\), then \(\psi(n) \to 0\).

3. If \(\lim |\phi(n)| = 0\), then \(\lim \phi(n) = 0\).

4. If \(\phi(n)\) tends to a limit or oscillates finitely, and \(|\psi(n)| \leq |\phi(n)|\) when \(n \geq n_{0}\), then \(\psi(n)\) tends to a limit or oscillates finitely.

5. If \(\phi(n)\) tends to \(+\infty\), or to \(-\infty\), or oscillates infinitely, and \[|\psi(n)| \geq |\phi(n)|\] when \(n \geq n_{0}\), then \(\psi(n)\) tends to \(+\infty\) or to \(-\infty\) or oscillates infinitely.

6. ‘If \(\phi(n)\) oscillates and, however great be \(n_{0}\), we can find values of \(n\) greater than \(n_{0}\) for which \(\psi(n) > \phi(n)\), and values of \(n\) greater than \(n_{0}\) for which \(\psi(n) < \phi(n)\), then \(\psi(n)\) oscillates’. Is this true? If not give an example to the contrary.

7. If \(\phi(n) \to l\) as \(n \to \infty\), then also \(\phi(n + p) \to l\), \(p\) being any fixed integer. [This follows at once from the definition. Similarly we see that if \(\phi(n)\) tends to \(+\infty\) or \(-\infty\) or oscillates so also does \(\phi(n + p)\).]

8. The same conclusions hold (except in the case of oscillation) if \(p\) varies with \(n\) but is always numerically less than a fixed positive integer \(N\); or if \(p\) varies with \(n\) in any way, so long as it is always positive.

9. Determine the least value of \(n_{0}\) for which it is true that \[(a)\ n^{2} + 2n > 999,999\quad (n \geq n_{0}),\qquad (b)\ n^{2} + 2n > 1,000,000\quad (n \geq n_{0}).\]

10. Determine the least value of \(n_{0}\) for which it is true that \[(a)\ n + (-1)^{n} > 1000\quad (n \geq n_{0}),\qquad (b)\ n + (-1)^{n} > 1,000,000\quad (n \geq n_{0}).\]

11. Determine the least value of \(n_{0}\) for which it is true that \[(a)\ n^{2} + 2n > \Delta\quad (n \geq n_{0}),\qquad (b)\ n + (-1)^{n} > \Delta\quad (n \geq n_{0}),\] \(\Delta\) being any positive number.

[(a) \(n_{0} = [\sqrt{\Delta + 1}]\): () \(n_{0} = 1 + [\Delta]\) or \(2 + [\Delta]\), according as \([\Delta]\) is odd or even, i.e. \(n_{0} = 1 + [\Delta] + \frac{1}{2} \{1 + (-1)^{[\Delta]}\}\).]

12. Determine the least value of \(n_{0}\) such that \[(a)\ n/(n^{2} + 1) < .0001,\qquad (b)\ (1/n) + \{(-1)^{n}/n^{2}\} < .000 01,\] when \(n \geq n_{0}\). [Let us take the latter case. In the first place \[(1/n) + \{(-1)^{n}/n^{2}\} \leq (n + 1)/n^{2},\] and it is easy to see that the least value of \(n_{0}\), such that \((n + 1)/n^{2} < .000 001\) when \(n \geq n_{0}\), is \(1,000,002\). But the inequality given is satisfied by \(n = 1,000,001\), and this is the value of \(n_{0}\) required.]


  1. See Bromwich’s Infinite Series, p. 485.↩︎

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