Infinite geometric series

When you reach for it

The sequence \displaystyle 12, 4, \frac{4}{3}, \frac{4}{9}, \ldots is geometric, with first term a_1 = 12 and common ratio \displaystyle r = \frac{1}{3}. It never ends, so the sum

can never be finished by adding one term after another. Yet the running totals behave very calmly. They are 12, then 16, then \displaystyle 17\frac{1}{3}, then \displaystyle 17\frac{7}{9}, and each one is closer to 18 than the last, without ever passing it. This lesson makes that precise: the infinite sum is 18, exactly.

Reach for this skill whenever the thing wanted is a total of infinitely many terms, each the same multiple of the one before. Such totals turn up in three common places. A repeating decimal such as 0.2\overline{4} = 0.2444\ldots is one, since its digits are an endless sum of shrinking pieces, and this skill turns it into the fraction \displaystyle \frac{11}{45}. Anything that repeats with a fixed fraction left over each time is another: a bungee jumper who springs back three quarters of the way after each drop, a medicine of which a quarter is still in the body when the next dose is taken. And in calculus functions are written as endless sums of powers of x, and the geometric series is the first and simplest of them.

An infinite geometric series is the sum of all the terms of a geometric sequence,

written in sigma notation as \sum_{k=1}^{\infty} a_1 r^{k-1}. The symbol \infty on top says that the index never stops.

The procedure

What the sum of infinitely many terms means

Nobody can carry out infinitely many additions, so the sum of an infinite series needs a definition. It is built from the finite sums that can be done. The partial sum S_n is the total of the first n terms, the same S_n as in arithmetic series and finite geometric series. A series has a sum when its partial sums settle down:

Definition. If the partial sums S_1, S_2, S_3, \ldots get as close as you like to one number S, and stay that close, then the series converges and S is its sum. If they do not approach any one number, the series diverges and has no sum.

The sum is therefore not the result of an endless process. It is the single number the finite totals close in on, and it is exact: the gap between S_n and S can be made smaller than any positive number by taking n large enough, and nothing else could be called the sum.

Why the partial sums of 12 + 4 + \cdots approach 18

The finite formula from Geometric series gives every partial sum at once. With a_1 = 12 and \displaystyle r = \frac{1}{3},

so \displaystyle S_n = 18 - \frac{18}{3^n}. The gap between S_n and 18 is \displaystyle \frac{18}{3^n}, which is 6, then 2, then \displaystyle \frac{2}{3}, then \displaystyle \frac{2}{9}, a third of the gap before each time. It is always positive, so no partial sum reaches 18; and it shrinks toward 0, so the partial sums get as close to 18 as you like. By the definition, the sum is 18.

Figure 1 shows the same thing as lengths. The first piece, 12, leaves a gap of 6 before 18. The next piece, 4, fills two thirds of that gap and leaves 2. The piece after, \displaystyle \frac{4}{3}, fills two thirds of what is left and leaves \displaystyle \frac{2}{3}. Every piece fills two thirds of the remaining gap, which is why the pieces can go on for ever without passing 18.

The formula

The same argument works for any ratio whose size is less than 1. The finite sum is

The first part does not depend on n at all. The second part contains r^n, and when |r| < 1 the powers r^n shrink toward 0 as n grows. (Why: if 0 < |r| < 1, write \displaystyle \frac{1}{|r|} = 1 + h with h > 0. Multiplying out (1 + h)^n gives 1 + nh plus terms that are all positive, so (1 + h)^n > nh and \displaystyle |r|^n < \frac{1}{nh}, which falls below any positive number once n is large. If r = 0 every power is already 0.) So the second part shrinks toward 0 and the partial sums approach the first part.

Sum of an infinite geometric series. If |r| < 1, the series a_1 + a_1 r + a_1 r^2 + \cdots converges, and its sum is

If |r| \ge 1 and a_1 \ne 0, the series diverges and has no sum.

The formula needs only two numbers. The first term a_1 is the first term actually in the sum. The ratio r is any term divided by the one before it, the same all the way along. There is no n, because there is no last term.

The split above also gives the gap after n terms exactly:

When r is positive this gap has the sign of a_1 throughout, so the partial sums approach from one side, as the sums of 12 + 4 + \cdots approach 18 from below. When r is negative, r^n changes sign at every step, and the partial sums land above and below S in turn, as in Example 2.

Why |r| \ge 1 gives no sum

If a series converges, its terms must shrink toward 0. The nth term is the difference of two partial sums, a_n = S_n - S_{n-1}, and if both S_n and S_{n-1} are close to S, their difference is close to 0. A geometric series with |r| \ge 1 fails this test, since every term has size |a_1 r^{n-1}| \ge |a_1|, which is never close to 0. So it diverges.

The four ways it can happen look different. When r > 1 the terms grow and so do the partial sums, past every number. When r = 1 every term is a_1 and S_n = n a_1, which also grows without end. When r = -1 the partial sums go a_1, 0, a_1, 0, \ldots for ever, never settling on one number. When r < -1 the terms alternate in sign and grow in size, and the partial sums swing further each time.

The formula \displaystyle \frac{a_1}{1 - r} still produces a number when |r| \ge 1 (and divides by 0 when r = 1), but that number is not the sum of anything. Check |r| < 1 first, every time.

Repeating decimals

A repeating decimal is an infinite geometric series in disguise, once the part that repeats is cut into its copies. Take 0.2\overline{4} = 0.24444\ldots. The 2 in the first place stands alone, and the 4s that follow are

a geometric series with first term \displaystyle \frac{4}{100} and ratio \displaystyle \frac{1}{10}, since each 4 is one place further right than the one before. Its sum is

and so

Dividing 11 by 45 gives 0.2444\ldots, which checks it. The same split works for any repeating decimal: the digits before the bar make a terminating decimal, and the repeating block makes a geometric series whose ratio is \displaystyle \frac{1}{10^p}, where p is the number of digits in the block. Since the ratio is less than 1 the series always converges, and its sum is always a fraction, which is the series view of why every repeating decimal is a rational number.

Worked examples

Boundaries

The ratio must satisfy |r| < 1. For |r| \ge 1 (and a_1 \ne 0) the series diverges, and \displaystyle \frac{a_1}{1 - r} is then a number that is not the sum of anything, such as the -4 in Example 1. At r = -1 the terms do not grow, yet there is still no sum, because the partial sums take two values for ever. The one trivial exception is a_1 = 0: every term is 0 and the sum is 0 whatever r is.

The series must be geometric. The formula needs the same ratio between every pair of neighboring terms, which is why the examples test two pairs. Shrinking terms on their own are not enough for a sum: \displaystyle 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots has terms that shrink toward 0 and still grows past every number, a fact taken up in Convergence tests. Terms that shrink toward 0 are needed for a sum, not sufficient.

The first term is the first term in the sum. In sigma notation it is the value at the lower limit of the index, as in Example 3, not the coefficient written in front.

The sum is a limit. It is the number the partial sums approach, not the end of a process that finishes. Within that definition it is exact, so 0.\overline{9} = 1 and 1.2\overline{9} = 1.3 are equalities, not approximations. No partial sum need equal S, and when r \ne 0 and a_1 \ne 0 none does, since the gap \displaystyle \frac{a_1 r^n}{1 - r} is never 0.

A model that rebounds for ever is still a model. In a setting such as Example 8 the infinite series gives a value that real events can approach but never complete, since a real jumper comes to rest after finitely many rebounds.

Where this goes next

The definition here, a sum as the number its partial sums approach, is the first appearance of a limit. What a limit is: graphs and tables studies limits on their own, and Infinite series and partial sums uses this definition for series of every kind, not only geometric ones. Most series have no tidy formula for S_n, so deciding whether they converge takes other tools: the The geometric series test packages this lesson’s result, and Convergence tests compares other series against geometric ones, since they are the series whose behavior is known exactly.

Example 7 points further still. A series such as 1 + 2x + 4x^2 + \cdots is a function of x written as an endless polynomial, and it equals \displaystyle \frac{1}{1 - 2x} on an interval. Writing functions this way is the idea behind Taylor and Maclaurin series, where the question of which x a series works for is settled at the ends of the interval by Power series and the interval of convergence.

Repeating decimals connect back to turning a fraction into a decimal by division: long division gives every fraction a terminating or repeating decimal, and the series argument here proves the reverse, that every repeating decimal names a fraction, so the two descriptions of the rational numbers match.

Exercises

Converge or diverge, and the sum

Decide whether each series converges. Give the sum of each one that does.

1. 50 + 40 + 32 + \cdots
2. \displaystyle 8 + 2 + \frac{1}{2} + \cdots
3. \displaystyle 7 - \frac{7}{2} + \frac{7}{4} - \cdots
4. \displaystyle 3 + 4 + \frac{16}{3} + \cdots
5. 5 - 5 + 5 - 5 + \cdots
6. 2.4 + 0.6 + 0.15 + \cdots

Sigma notation

7. Find \displaystyle \sum_{k=1}^{\infty} 10\left(\frac{1}{5}\right)^{k}.
8. Find \displaystyle \sum_{k=0}^{\infty} 6\left(-\frac{3}{5}\right)^{k}.
9. Does \sum_{k=1}^{\infty} 2(1.05)^{k} have a sum? Explain.

Repeating decimals

Write each decimal as a fraction in lowest terms by treating it as an infinite geometric series.

10. 0.\overline{513}
11. 0.6\overline{1}
12. 3.\overline{12}
13. 0.0\overline{75}
14. 2.3\overline{9}

Working backwards

15. An infinite geometric series has first term 6 and sum 24. Find its ratio.
16. An infinite geometric series has ratio \displaystyle -\frac{1}{4} and sum 10. Find its first term.
17. An infinite geometric series has second term 6 and sum 25. Find every possible first term and ratio.
18. For which values of x does \displaystyle \sum_{k=0}^{\infty} \left(\frac{x - 1}{2}\right)^{k} converge, and what is its sum?

In a setting

19. A pendulum is let go, and its first swing carries the bob along an arc of 25 centimeters. Each swing after that is \displaystyle \frac{3}{5} as long as the one before. If it swung for ever, how far would the bob travel in total?
20. A patient takes 200 milligrams of a medicine every morning, and by the next morning a quarter of the amount in the body is left. After many days, how much is in the body just after a dose? Give the exact value and one decimal place.
21. How many terms of 48 + 12 + 3 + \cdots are needed before the partial sum is within 0.001 of the sum of the whole series?

Reasoning

22. The formula \displaystyle \frac{a_1}{1 - r} applied to 1 + 2 + 4 + 8 + \cdots gives -1. Explain why the series has no sum, and what went wrong.
23. Show that \displaystyle S - S_n = \frac{a_1 r^n}{1 - r} for |r| < 1, and use it to explain why, when a_1 > 0 and 0 < r < 1, every partial sum is less than S.
24. The terms of 1 - 1 + 1 - 1 + \cdots do not grow. Explain why the series still has no sum.
25. A decimal 0.\overline{B} repeats a block B of p digits from the decimal point on. Prove that it equals \displaystyle \frac{B}{10^p - 1}, and explain why this shows every such decimal is a rational number.