78. The infinite geometrical series

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
$$\begin{array}{rl}\frac{2}{10}+\frac{1}{{10}^2}+\frac{7}{{10}^3}+\frac{1}{{10}^4}+\frac{3}{{10}^5}+\frac{1}{{10}^6}+\frac{3}{{10}^7}+\dots & =\frac{217}{1000}+\frac{\frac{13}{{10}^5}}{1–\frac{1}{{10}^2}}\\ & =\frac{2687}{12,375}.\end{array}$$
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 $$.\bar{9}=\frac{9}{10}+\frac{9}{{10}^2}+\frac{9}{{10}^3}+\cdots =1,$$ 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, $.217=.216\bar{9}$. 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 $.\bar{9}$ 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 $a_r\ge b_r+1>b_r.b_{r+1}{b}_{r+2}\dots$ (unless $b_{r+1}$, $b_{r+2}$, … are all $9$’s), and so $$.a_1{a}_2\dots a_r{a}_{r+1}\cdots >.b_1{b}_2\dots b_r{b}_{r+1}\dots .$$ 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 $$0,1/10;1/10,2/10;\dots ;9/10,1.$$ 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 $.1010010001000010\dots$ and $.2020020002000020\dots$, 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 $.11101010001010\dots$, 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

1. The series $r^m+r^{m+1}+\dots$ is convergent if $-1<r<1$, and its sum is $1/(1–r)–1–r–\cdots –r^{m-1}$ (§ 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 $+\mathrm{\infty }$ or to $-\mathrm{\infty }$ according to the sign of $b$, while if $b=0$ it diverges to $+\mathrm{\infty }$ or $-\mathrm{\infty }$ 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 $-1<r\le 1$. 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 $1+r+r^2+\cdots =s$ 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 $$\begin{array}{rlrl}& r–\frac{r}{1+r}+\frac{r}{(1+r{)}^2}–\dots ,& & r+\frac{r}{1–r}+\frac{r}{(1–r{)}^2}+\dots ,\\ & 1–\frac{r}{1+r}+{\left(\frac{r}{1+r}\right)}^2–\dots ,& & 1+\frac{r}{1–r}+{\left(\frac{r}{1–r}\right)}^2+\dots .\end{array}$$

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

11. If $0\le a_n\le 1$ then the series $a_0+a_{1}r+a_2{r}^2+\dots$ is convergent for $0\le r<1$, 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 $0\le r\le 1$, and its sum is not greater than the lesser of $a_0+a_1+a_2+\dots$ and $1/(1–r)$.

13. The series $$1+\frac{1}{1}+\frac{1}{1\cdot 2}+\frac{1}{1\cdot 2\cdot 3}+\dots$$ is convergent. [For $1/(1\cdot 2\dots n)\le 1/2^{n-1}$.]

The series $$1+\frac{1}{1\cdot 2}+\frac{1}{1\cdot 2\cdot 3\cdot 4}+\dots ,\frac{1}{1}+\frac{1}{1\cdot 2\cdot 3}+\frac{1}{1\cdot 2\cdot 3\cdot 4\cdot 5}+\dots$$ 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 $+\mathrm{\infty }$.

[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 $n\to \mathrm{\infty }$.

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.