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Geometric series

A geometric sequence multiplies by a constant ratio r at each step; the terms increase in size when |r| > 1 and tend to zero when |r| < 1. Multiplying the sum by r and subtracting proves the sum formula. An infinite geometric series converges when |r| < 1, and logarithms give the number of terms needed to reach a stated total.

Year 12-13EDEXCEL 9MA0 4.5, 4.6

Builds on Arithmetic series and Logarithms and their laws.

IN THIS TOPIC

  • Use un = arn−1, and recover a and r from two given terms.
  • Prove the finite sum formula by the subtraction trick, and use it.
  • Say when a series converges, and find the sum to infinity.
  • Answer how-many-terms and how-many-years questions with logarithms.
  • Apply geometric series to modelling, compound growth included.

COMMON MISCONCEPTION

A sum that never ends must be infinite.

An infinite geometric series has a finite sum whenever |r| < 1, because the terms then decrease in size geometrically. For 8 + 4 + 2 + … the ratio is ½, and the partial sums increase towards 8/(1 − ½) = 16 without reaching it. State the condition |r| < 1 before using S∞ = a/(1 − r).

Ratio, term, sum

A geometric sequence multiplies by a fixed common ratio r at each step, so its nth term is

un=arn−1u_{n} = ar^{n−1}NOT IN THE BOOKLET: LEARN IT

and the sum of the first n terms, printed in the booklet, is

Sn=a(1−rn)1−rS_{n} = \frac{a(1 − r^{n})}{1 − r}IN THE FORMULAE BOOKLET

Edexcel 9MA0 4.5 requires this proof. Multiply Sn by r and subtract the two lines: all but two terms cancel, leaving Sn − rSn = a − arn, and dividing by 1 − r, which requires r ≠ 1, gives the formula. ∎

WORKED EXAMPLE

A doubling series, summed

Find the sum of the first 10 terms of 3 + 6 + 12 + …

Here a = 3 and r = 2, so S10 = 3(210 − 1)/(2 − 1) = 3 × 1023 = 3069.

With r > 1 it is tidier to flip both brackets and write a(rn − 1)/(r − 1), which keeps every quantity positive.

The last term alone is 3 × 29 = 1536, more than half the total. Where |r| > 1 the later terms dominate the sum, which is useful when checking whether a total is of the right size.

The infinite sum

Where |r| < 1, rn tends to zero as n increases, so the term arn in the sum formula tends to zero and the sum tends to a finite limit,

S∞=a1−rS_{∞} = \frac{a}{1 − r}IN THE FORMULAE BOOKLET

An infinite sum has a finite value provided the terms decrease in size geometrically, that is provided |r| < 1.

A long rectangle of length 16 filled left to right by colour-alternating segments of computed lengths 8, 4, 2, 1 and onward, each half its neighbour: the segments crowd the right-hand end without ever crossing it, which is what a sum to infinity of 16 means.
FIG. 18 + 4 + 2 + 1 + … drawn along a line of length 16. The partial sums increase towards 16 without reaching it, and 16 is the sum to infinity.

GUIDED PRACTICE

Convergent, and summed

For the series 8 + 4 + 2 + …, explain why the sum to infinity exists and find it, before opening the working.

Show the working

The ratio is r = ½. Since |½| < 1, the series converges.

S∞ = 8/(1 − ½) = 16.

State the condition |r| < 1 before using the formula: the sum to infinity exists only when it holds.

Logs answer how long

A question asking how many terms, or how many years of compound growth, puts the unknown in an exponent, which is solved by taking logarithms of both sides.

Twenty-one computed dots of a 2000 pound investment compounding at 4 percent, bending upward across a dashed line at 4000: year 17 sits just below the line, and the first amber dot above it is year 18, exactly where the logarithm said the doubling would land.
FIG. 2£2000 at 4% compound interest. The values are geometric with r = 1.04, and the first value above £4000 is at year 18, which agrees with n > ln 2/ln 1.04.

INDEPENDENT PRACTICE

Money doubling

£2000 is invested at 4% compound interest per year. Show that the value after n years is 2000 × 1.04n, and find the first year in which the money has more than doubled.

Show the working

Each year multiplies the value by 1.04, so after n years it is 2000 × 1.04n, geometric growth with ratio 1.04.

Doubling needs 1.04n > 2. Taking logs gives n > ln 2/ln 1.04 = 17.67.

The first whole year past that is year 18, where the value is £4052. Year 17 gives £3896 and falls short.

Round up, then give the values for the years either side so that the crossing is shown.

ASSESSMENT FOCUS

  • Write a and r down before using a formula: r is the ratio of consecutive terms, the second divided by the first.
  • To prove the sum formula, multiply Sn by r, subtract the two lines, and show the cancellation.
  • State |r| < 1 before using the sum to infinity: it is the condition for the series to converge.
  • For r > 1, use the equivalent form a(rn − 1)/(r − 1), in which both brackets are positive.
  • Recovering r from two non-adjacent terms means dividing, not subtracting. If u6/u3 = 8 then r3 = 8 and r = 2.
  • A question asking for a number of terms ends with a logarithm and a rounding up. Give the neighbouring values so that the crossing is shown.

CHECK YOURSELF

For the series 5 + 4 + 3.2 + …, find the sum to infinity, and the sum of the first 10 terms to 4 significant figures.

Show a hint

The ratio is 0.8; then apply both formulae.

Show the answer

r = 4/5 = 0.8, and |0.8| < 1, so S∞ = 5/(1 − 0.8) = 25.

S10 = 5(1 − 0.810)/0.2 = 22.32 to 4 significant figures.

Ten terms give about 89% of the sum to infinity, since 0.810 ≈ 0.107.

Each term is r times the previous one, and subtracting rSn from Sn cancels all but two terms, which proves the sum formula for r ≠ 1.

The series converges when |r| < 1, with sum to infinity a/(1 − r); take logarithms to find the number of terms needed to reach a stated total.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

Question workbook

PDF · 55 KB · print edition · white background

Question workbook

PDF · 55 KB · screen edition · navy

Answer book

PDF · 67 KB · fully worked · navy
15 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the geometric series questions page.

USED LATER IN

CHECK YOUR PROGRESS

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  • Use un = arn−1, and recover a and r from two given terms.
  • Prove the finite sum formula by the subtraction trick, and use it.
  • Say when a series converges, and find the sum to infinity.
  • Answer how-many-terms and how-many-years questions with logarithms.
  • Apply geometric series to modelling, compound growth included.

Open the full revision checklist to see every objective in the course in one place.