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.↩︎