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 \(r < -1\).

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}

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.

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