We shall now consider the ‘geometrical’ series, whose general term is $$u_{n} = r^{n-1}$$. In this case $s_{n} = 1 + r + r^{2} + \dots + r^{n-1} = (1 – r^{n})/(1 – r),$ except in the special case in which $$r = 1$$, when $s_{n} = 1 + 1 + \dots + 1 = n.$ In the last case $$s_{n} \to +\infty$$. In the general case $$s_{n}$$ will tend to a limit if and only if $$r^{n}$$ does so. Referring to the results of § 72 we see that

the series $$1 + r + r^{2} + \dots$$ is convergent and has the sum $$1/(1 – r)$$ if and only if $$-1 < r < 1$$.

If $$r \geq 1$$, then $$s_{n} \geq n$$, and so $$s_{n} \to +\infty$$; the series diverges to $$+\infty$$. If $$r = -1$$, then $$s_{n} = 1$$ or $$s_{n} = 0$$ according as $$n$$ is odd or even: $$s_{n}$$ oscillates finitely. If $$r < -1$$, then $$s_{n}$$ oscillates infinitely. Thus, to sum up,

the series $$1 + r + r^{2} + \dots$$ diverges to $$+\infty$$ if $$r \geq 1$$, converges to $$1/(1 – r)$$ if $$-1 < r < 1$$, oscillates finitely if $$r = -1$$, and oscillates infinitely if $$r < -1$$.

Example XXIX

1. Recurring decimals. The commonest example of an infinite geometric series is given by an ordinary recurring decimal. Consider, for example, the decimal $$.217{\overline{13}}$$. This stands, according to the ordinary rules of arithmetic, for
\begin{align*}
\frac{2}{10} + \frac{1}{10^{2}} + \frac{7}{10^{3}} + \frac{1}{10^{4}} + \frac{3}{10^{5}} + \frac{1}{10^{6}} + \frac{3}{10^{7}} + \dots &= \frac{217}{1000} +\frac{ \frac{13}{10^{5}} }{1 – \frac{1}{10^{2}}}\\
&= \frac{2687}{12,375}.
\end{align*}
The reader should consider where and how any of the general theorems of § 77 have been used in this reduction.

2. Show that in general $.a_{1}a_{2}\dots a_{m} {\overline{\alpha_{1}\alpha_{2}\dots \alpha_{n}}} = \frac{a_{1}a_{2}\dots a_{m}\alpha_{1}\dots \alpha_{n} – a_{1}a_{2}\dots a_{n}} {99\dots 900\dots 0},$ the denominator containing $$n$$ $$9$$’s and $$m$$ $$0$$’s.

3. Show that a pure recurring decimal is always equal to a proper fraction whose denominator does not contain $$2$$ or $$5$$ as a factor.

4. A decimal with $$m$$ non-recurring and $$n$$ recurring decimal figures is equal to a proper fraction whose denominator is divisible by $$2^{m}$$ or $$5^{m}$$ but by no higher power of either.

5. The converses of Exs. 3, 4 are also true. Let $$r = p/q$$, and suppose first that $$q$$ is prime to $$10$$. If we divide all powers of $$10$$ by $$q$$ we can obtain at most $$q$$ different remainders. It is therefore possible to find two numbers $$n_{1}$$ and $$n_{2}$$, where $${n_{1} > n_{2}}$$, such that $$10^{n_{1}}$$ and $$10^{n_{2}}$$ give the same remainder. Hence $$10^{n_{1}} – 10^{n_{2}} = 10^{n_{2}}(10^{n_{1}-n_{2}} – 1)$$ is divisible by $$q$$, and so $$10^{n} – 1$$, where $$n = n_{1} – n_{2}$$, is divisible by $$q$$. Hence $$r$$ may be expressed in the form $$P/(10^{n} – 1)$$, or in the form $\frac{P}{10^{n}} + \frac{P}{10^{2n}} + \dots,$ i.e. as a pure recurring decimal with $$n$$ figures. If on the other hand $$q = 2^{\alpha}5^{\beta}Q$$, where $$Q$$ is prime to $$10$$, and $$m$$ is the greater of $$\alpha$$ and $$\beta$$, then $$10^{m}r$$ has a denominator prime to $$10$$, and is therefore expressible as the sum of an integer and a pure recurring decimal. But this is not true of $$10^{\mu}r$$, for any value of $$\mu$$ less than $$m$$; hence the decimal for $$r$$ has exactly $$m$$ non-recurring figures.

6. To the results of Exs. 2–5 we must add that of Ex. I. 3. Finally, if we observe that $.{\overline{9}} = \frac{9}{10} + \frac{9}{10^{2}} + \frac{9}{10^{3}} + \dots = 1,$ we see that every terminating decimal can also be expressed as a mixed recurring decimal whose recurring part is composed entirely of $$9$$’s. For example, $$.217 = .216{\overline{9}}$$. Thus every proper fraction can be expressed as a recurring decimal, and conversely.

7. Decimals in general. The expression of irrational numbers as non-recurring decimals. Any decimal, whether recurring or not, corresponds to a definite number between $$0$$ and $$1$$. For the decimal $$.a_{1}a_{2}a_{3}a_{4}\dots$$ stands for the series $\frac{a_{1}}{10} + \frac{a_{2}}{10^{2}} + \frac{a_{3}}{10^{3}} + \dots.$ Since all the digits $$a_{r}$$ are positive, the sum $$s_{n}$$ of the first $$n$$ terms of this series increases with $$n$$, and it is certainly not greater than $$.{\overline{9}}$$ or $$1$$. Hence $$s_{n}$$ tends to a limit between $$0$$ and $$1$$.

Moreover no two decimals can correspond to the same number (except in the special case noticed in Ex. 6). For suppose that $$.a_{1}a_{2}a_{3} \dots$$, $$.b_{1}b_{2}b_{3} \dots$$ are two decimals which agree as far as the figures $$a_{r-1}$$, $$b_{r-1}$$, while $$a_{r} > b_{r}$$. Then $$a_{r}\geq b_{r} + 1 > b_{r}.b_{r+1}b_{r+2} \dots$$ (unless $$b_{r+1}$$, $$b_{r+2}$$, … are all $$9$$’s), and so $.a_{1}a_{2} \dots a_{r}a_{r+1} \dots > .b_{1}b_{2} \dots b_{r}b_{r+1} \dots.$ It follows that the expression of a rational fraction as a recurring decimal (Exs. 2–6) is unique. It also follows that every decimal which does not recur represents some irrational number between $$0$$ and $$1$$. Conversely, any such number can be expressed as such a decimal. For it must lie in one of the intervals $0,\ 1/10;\quad 1/10,\ 2/10;\ \dots;\quad 9/10,\ 1.$ If it lies between $$r/10$$ and $$(r + 1)/10$$, then the first figure is $$r$$. By subdividing this interval into $$10$$ parts we can determine the second figure; and so on. But (Exs. 3, 4) the decimal cannot recur. Thus, for example, the decimal $$1.414\dots$$, obtained by the ordinary process for the extraction of $$\sqrt{2}$$, cannot recur.

8. The decimals $$.101\ 001\ 000\ 100\ 001\ 0\dots$$ and $$.202\ 002\ 000\ 200\ 002\ 0\dots$$, in which the number of zeros between two $$1$$’s or $$2$$’s increases by one at each stage, represent irrational numbers.

9. The decimal $$.111\ 010\ 100\ 010\ 10\dots$$, in which the $$n$$th figure is $$1$$ if $$n$$ is prime, and zero otherwise, represents an irrational number. [Since the number of primes is infinite the decimal does not terminate. Nor can it recur: for if it did we could determine $$m$$ and $$p$$ so that $$m$$, $$m + p$$, $$m + 2p$$, $$m + 3p$$, … are all prime numbers; and this is absurd, since the series includes $$m + mp$$.]1

Example XXX

1. The series $$r^{m} + r^{m+1} + \dots$$ is convergent if $$-1 < r < 1$$, and its sum is $$1/(1 – r) – 1 – r – \dots – r^{m-1}$$ (§ 77, (2)).

2. The series $$r^{m} + r^{m+1} + \dots$$ is convergent if $$-1 < r < 1$$, and its sum is $$r^{m}/(1 – r)$$ (§ 77, (4)). Verify that the results of Exs. 1 and 2 are in agreement.

3. Prove that the series $$1 + 2r + 2r^{2} + \dots$$ is convergent, and that its sum is $$(1 + r)/(1 – r)$$, ($$\alpha$$) by writing it in the form $$-1 + 2(1 + r + r^{2} + \dots)$$, ($$\beta$$) by writing it in the form $$1 + 2(r + r^{2} + \dots)$$, ($$\gamma$$) by adding the two series $$1 + r + r^{2} + \dots$$, $$r + r^{2} + \dots$$. In each case mention which of the theorems of § 77 are used in your proof.

4. Prove that the ‘arithmetic’ series $a + (a + b) + (a + 2b) + \dots$ is always divergent, unless both $$a$$ and $$b$$ are zero. Show that, if $$b$$ is not zero, the series diverges to $$+\infty$$ or to $$-\infty$$ according to the sign of $$b$$, while if $$b = 0$$ it diverges to $$+\infty$$ or $$-\infty$$ according to the sign of $$a$$.

5. What is the sum of the series $(1 – r) + (r – r^{2}) + (r^{2} – r^{3}) + \dots$ when the series is convergent? [The series converges only if $$-1 < r \leq 1$$. Its sum is $$1$$, except when $$r = 1$$, when its sum is $$0$$.]

6. Sum the series $r^{2} + \frac{r^{2}}{1 + r^{2}} + \frac{r^{2}}{(1 + r^{2})^{2}} + \dots.$ [The series is always convergent. Its sum is $$1 + r^{2}$$, except when $$r = 0$$, when its sum is $$0$$.]

7. If we assume that $$1 + r + r^{2} + \dots$$ is convergent then we can prove that its sum is $$1/(1 – r)$$ by means of § 77, (1) and (4). For if $$1 + r + r^{2} + \dots = s$$ then $s = 1 + r(1 + r^{2} + \dots) = 1 + rs.$

8. Sum the series $r + \frac{r}{1 + r} + \frac{r}{(1 + r)^{2}} + \dots$ when it is convergent. [The series is convergent if $$-1 < 1/(1 + r) < 1$$, i.e. if $$r < -2$$ or if $$r > 0$$, and its sum is $$1 + r$$. It is also convergent when $$r = 0$$, when its sum is $$0$$.]

9. Answer the same question for the series \begin{aligned} & r – \frac{r}{1 + r} + \frac{r}{(1 + r)^{2}} – \dots, && r + \frac{r}{1 – r} + \frac{r}{(1 – r)^{2}} + \dots,\\ & 1 – \frac{r}{1 + r} + \left(\frac{r}{1 + r}\right)^{2} – \dots, && 1 + \frac{r}{1 – r} + \left(\frac{r}{1 – r}\right)^{2} + \dots.\end{aligned}

10. Consider the convergence of the series \begin{aligned} & (1 + r) + (r^{2} + r^{3}) + \dots, && (1 + r + r^{2}) + (r^{3} + r^{4} + r^{5}) + \dots,\\ & 1 – 2r + r^{2} + r^{3} – 2r^{4} + r^{5} + \dots, && (1 – 2r + r^{2}) + (r^{3} – 2r^{4} + r^{5}) + \dots,\end{aligned} and find their sums when they are convergent.

11. If $$0 \leq a_{n} \leq 1$$ then the series $$a_{0} + a_{1}r + a_{2}r^{2} + \dots$$ is convergent for $$0 \leq r < 1$$, and its sum is not greater than $$1/(1 – r)$$.

12. If in addition the series $$a_{0} + a_{1} + a_{2} + \dots$$ is convergent, then the series $$a_{0} + a_{1}r + a_{2}r^{2} + \dots$$ is convergent for $$0 \leq r \leq 1$$, and its sum is not greater than the lesser of $$a_{0} + a_{1} + a_{2} + \dots$$ and $$1/(1 – r)$$.

13. The series $1 + \frac{1}{1} + \frac{1}{1\cdot2} + \frac{1}{1\cdot2\cdot3} + \dots$ is convergent. [For $$1/(1\cdot2 \dots n) \leq 1/2^{n-1}$$.]

The series $1 + \frac{1}{1\cdot2} + \frac{1}{1\cdot2\cdot3\cdot4} + \dots,\quad \frac{1}{1} + \frac{1}{1\cdot2\cdot3} + \frac{1}{1\cdot2\cdot3\cdot4\cdot5} + \dots$ are convergent.

15. The general harmonic series $\frac{1}{a} + \frac{1}{a + b} + \frac{1}{a + 2b} + \dots,$ where $$a$$ and $$b$$ are positive, diverges to $$+\infty$$.

[For $$u_{n} = 1/(a + nb) > 1/\{n(a + b)\}$$. Now compare with $$1 + \frac{1}{2} + \frac{1}{3} + \dots$$.]

16. Show that the series $(u_{0} – u_{1}) + (u_{1} – u_{2}) + (u_{2} – u_{3}) + \dots$ is convergent if and only if $$u_{n}$$ tends to a limit as $$n \to \infty$$.

17. If $$u_{1} + u_{2} + u_{3} + \dots$$ is divergent then so is any series formed by grouping the terms in brackets in any way to form new single terms.

18. Any series, formed by taking a selection of the terms of a convergent series of positive terms, is itself convergent.

1. All the results of Exs. XXIX may be extended, with suitable modifications, to decimals in any scale of notation. For a fuller discussion see Bromwich, Infinite Series, Appendix I.↩︎