The sum of an infinite geometric sequence

Higher Level

A takraw ball dropped from 2 m rebounds to 0.8 of its height each time. The bounces never stop. Count them and watch the total distance.

0 bounces

2.000next bounce, m

2.000distance so far

18limit

Every bounce is shorter. None of them is the last.

The formula and its condition

S_∞ = u₁1 − r, and it exists only when |r| < 1. If the terms do not shrink towards nothing, there is nothing for the running total to settle on, and the formula produces a number that means nothing at all.

a = 10, r = 0.6→ S_∞ = 100.4 = 25 S₁₀= 24.8488 S₂₀= 24.9991 closing in, never arriving r = 1.2→ no sum, the terms grow

The bouncing ball

The usual trap is counting the distance once. The ball goes down 2 m, then every rebound is travelled twice, up and then down again.

first drop= 2 rebounds= 2 × (1.6 + 1.28 + 1.024 + …) that bracket= 1.61 − 0.8 = 8 total= 2 + 2(8) = 18 m

0.999… = 1 by the same formula. It is 0.9 + 0.09 + 0.009 + …, a geometric series with a = 0.9 and r = 0.1, so the sum is 0.9/0.9 = 1 exactly. Not nearly 1. Exactly 1.

Worked example

A pendulum swings through 40 cm on its first swing. Each swing after that is 0.85 of the one before.

Find the total distance the bob travels before it stops.

u₁ = 40, r = 0.85Each swing is a fixed fraction of the last, so it is geometric, not arithmetic.
|r| = 0.85 < 1, so a sum existsState this BEFORE using the formula. Questions are set where it fails, precisely to see whether you check.
S_∞ = 40 ÷ (1 − 0.85)a over 1 minus r. Note 1 − r is 0.15, a small number, so the total will be large.
= 40 ÷ 0.15Dividing by a small number multiplies.
= 266.7 cm, about 2.67 mNearly seven times the first swing, from a bob that never quite stops. The swings continue for ever in the model and the total is finite, which is the whole idea of this sub-topic.

Your turn

1. a = 10 and r = 0.6. Find the sum to infinity.

2. A takraw ball is dropped from 2 m and rebounds to 0.8 of its height. Find the total distance travelled.

3. A geometric sequence has r = 1.2. Its sum to infinity:

Is u₁/(1 − 1.2) = −5u₁. Does not exist, because |r| is not less than 1. Is zero. Equals u₁.

Where the marks go

Stating |r| < 1 before using the formula. Questions are set where it fails, precisely to see whether you check.

Doubling the rebounds and not the first drop, on a bouncing ball question.

Giving an exact value when one exists. 18 is the answer; 17.9 is a partial sum.

Come back to this

These three are deliberately not all about this page. In an examination the hardest step is deciding which method applies, and questions that arrive straight after the method never make you decide. Answer from memory before you reveal anything.

Come back to these in a week, and again in a month.

Which method?

Multiply 2cis(π/3) by 3cis(π/6).

Polar form: multiply the moduli, add the arguments. 6cis(π/2) = 6i. One line, against four terms in Cartesian form.
Which method?

A sequence has r = 1.2. Find its sum to infinity.

It does not exist. |r| is not less than 1, so the terms grow. The formula would return a negative number for a sum of positive terms, which is the giveaway.
Which method?

Find the upper bound of a length given as 12 m to the nearest metre.

Bounds: 12.5 m, and note it is strictly less than, because 12.5 would round up.

The same idea elsewhere

5.1, limits and gradient: A limit is the same idea: approaching a value without reaching it.
4.19, transition matrices: A steady state is reached by a geometric decay of exactly this kind.

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