Unit 7, lesson 94

Geometric Series

How to add a geometric series exactly, why it converges precisely when the absolute value of the ratio is less than one, and how this turns repeating decimals into fractions.

  • Theory part 187
  • Practice part 188
  • Infinite Series

A ball bounces back to two thirds of its previous height every time. The heights form a list in which each number is the one before it times the same factor, and the total distance the ball travels comes from adding up that list. Series of this kind are the one family where you can write down the sum exactly, which makes them the yardstick that many other series get measured against.

Key terms

Common ratio

Definition A geometric sequence is a sequence of the form a,ar,ar2,ar3,…a, ar, ar^2, ar^3, \ldots with first term a≠0a \ne 0 and a fixed real number rr, called the common ratio. Its terms are an=arna_n = ar^n for n≥0n \ge 0, and when every term is nonzero

r=an+1anfor every n.r = \frac{a_{n+1}}{a_n} \quad \text{for every } n.

In plain words Each term is the previous one multiplied by the same number rr. Divide any term by the one before it and you get rr.

Example In 12,−6,3,−32,…12, -6, 3, -\frac{3}{2}, \ldots each term is the previous one times −12-\frac{1}{2}, so a=12a = 12 and r=−12r = -\frac{1}{2}. A negative ratio makes the signs alternate.

Finite geometric sums

Definition For a≠0a \ne 0 and any integer N≥1N \ge 1,

∑k=0N−1ark=a+ar+⋯+arN−1={a(1−rN)1−r,r≠1,Na,r=1.\sum_{k=0}^{N-1} ar^k = a + ar + \cdots + ar^{N-1} = \begin{cases} \dfrac{a(1 - r^N)}{1 - r}, & r \ne 1, \\[2mm] Na, & r = 1. \end{cases}

In plain words Call the sum ss. Multiply it by rr and subtract. Almost everything cancels, leaving s−rs=a−arNs - rs = a - ar^N, and you divide by 1−r1 - r.

Example The sum 3+6+12+⋯+3⋅293 + 6 + 12 + \cdots + 3 \cdot 2^9 has a=3a = 3, r=2r = 2 and N=10N = 10 terms, so it equals

3(1−210)1−2=3(1024−1)=3069.\frac{3(1 - 2^{10})}{1 - 2} = 3(1024 - 1) = 3069.

Infinite geometric series

Definition Let a≠0a \ne 0. The geometric series

∑n=0∞arn=a+ar+ar2+⋯\sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + \cdots

converges if and only if ∣r∣<1\lvert r \rvert < 1, and in that case its sum is

∑n=0∞arn=a1−r.\sum_{n=0}^{\infty} ar^n = \frac{a}{1 - r}.

In plain words For r≠1r \ne 1 the partial sums are

sN=a(1−rN)1−r.s_N = \frac{a(1 - r^N)}{1 - r}.

When ∣r∣<1\lvert r \rvert < 1 the power rNr^N shrinks to 00, so the sum is the first term divided by one minus the ratio.

Example For 12−6+3−32+⋯12 - 6 + 3 - \frac{3}{2} + \cdots we have ∣r∣=12<1\lvert r \rvert = \frac{1}{2} < 1, so the sum is

121−(−12)=123/2=8.\frac{12}{1 - \left(-\frac{1}{2}\right)} = \frac{12}{3/2} = 8.

Watch the starting index. The series ∑n=1∞(23)n\sum_{n=1}^{\infty} \left(\frac{2}{3}\right)^n has first term 23\frac{2}{3}, so its sum is

2/31−2/3=2,\frac{2/3}{1 - 2/3} = 2,

not 33.

The endpoints and beyond

Definition Let a≠0a \ne 0. If ∣r∣≥1\lvert r \rvert \ge 1, the geometric series ∑n=0∞arn\sum_{n=0}^{\infty} ar^n diverges. For r=1r = 1 the partial sums are NaNa, which tend to ∞\infty if a>0a > 0 and to −∞-\infty if a<0a < 0. For r=−1r = -1 the partial sums alternate between aa and 00. For ∣r∣>1\lvert r \rvert > 1 the terms satisfy ∣arn∣→∞\lvert ar^n \rvert \to \infty, so the partial sums cannot settle.

In plain words The boundary cases r=1r = 1 and r=−1r = -1 both fail, for different reasons. One runs away and the other keeps flipping.

Example With a=1a = 1 and r=−1r = -1 the series is 1−1+1−1+⋯1 - 1 + 1 - 1 + \cdots.

NN1122334455
sNs_N1100110011

The partial sums never close in on one value, so the series diverges, even though the formula a1−r\frac{a}{1 - r} would give 12\frac{1}{2}. The formula only holds for ∣r∣<1\lvert r \rvert < 1.

Repeating decimals

Definition A repeating decimal is a geometric series. If a block of kk digits forming the integer dd repeats forever right after the decimal point, then

0.d‾=d10k+d102k+⋯=d/10k1−10−k=d10k−1.0.\overline{d} = \frac{d}{10^k} + \frac{d}{10^{2k}} + \cdots = \frac{d/10^k}{1 - 10^{-k}} = \frac{d}{10^k - 1}.

In plain words Each repetition is the previous one shifted kk places to the right, which means multiplied by 10−k10^{-k}. So the decimal is a geometric series with ratio 10−k10^{-k}.

Example For 0.272727…0.272727\ldots the block is 2727 with k=2k = 2, so it equals 2799=311\frac{27}{99} = \frac{3}{11}. When the repeat starts later, split off the part before it. The repeating part of 1.2333…1.2333\ldots is a geometric series with first term 3100\frac{3}{100} and ratio 110\frac{1}{10}, so

1.2333…=1.2+3/1001−1/10=1.2+130=3730.1.2333\ldots = 1.2 + \frac{3/100}{1 - 1/10} = 1.2 + \frac{1}{30} = \frac{37}{30}.

The same rule shows 0.999…=99=10.999\ldots = \frac{9}{9} = 1.

Check yourself

  1. Find ∑n=1∞2⋅3n5n\sum_{n=1}^{\infty} \frac{2 \cdot 3^n}{5^n}.
Show the answer

It is geometric with first term 65\frac{6}{5} and ratio 35\frac{3}{5}, so the sum is

6/51−3/5=6/52/5=3.\frac{6/5}{1 - 3/5} = \frac{6/5}{2/5} = 3.
  1. Find ∑n=0∞(−2)n3n+1\sum_{n=0}^{\infty} \frac{(-2)^n}{3^{n+1}}.
Show the answer

The first term is 13\frac{1}{3} and the ratio is −23-\frac{2}{3}, with ∣r∣<1\lvert r \rvert < 1. The sum is

1/31−(−2/3)=1/35/3=15.\frac{1/3}{1 - (-2/3)} = \frac{1/3}{5/3} = \frac{1}{5}.
  1. For which xx does ∑n=0∞(x2)n\sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^n converge, and to what?
Show the answer

It is geometric with first term 11 and ratio x2\frac{x}{2}. It converges exactly when ∣x2∣<1\left\lvert \frac{x}{2} \right\rvert < 1, that is −2<x<2-2 < x < 2, and then the sum is

11−x/2=22−x.\frac{1}{1 - x/2} = \frac{2}{2 - x}.

How this lesson connects

It also uses geometric progressions.

Used later in