Maths › Sequences and series › Geometric series

Geometric series

Multiply by the same ratio each step and growth turns explosive, or decay turns endless. Geometric series sum by a telescoping trick, infinite ones can still total something finite when the ratio is small, and logarithms answer every how-long question compound growth can pose.

Year 12-13EDEXCEL 9MA0 4.5, 4.6

Builds on Arithmetic series and Logarithms and their laws.

IN THIS TOPIC

  • Use the nth-term formula arn−1, and prove and use the finite sum formula.
  • Recognise convergence and use the sum to infinity when |r| < 1.
  • Solve how-many-terms questions with logarithms, in series and compound-growth models.

WHAT YOU PROBABLY THINK

A sum that never ends must be infinite.

Ratio, term, sum

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

un = arn−1NOT IN THE BOOKLET — LEARN IT

and the finite sum, in the booklet, is

Sn = a(1 − rn)1 − rIN THE FORMULAE BOOKLET

with a proof the specification requires: multiply Sn by r and subtract. Every term but two cancels, leaving Sn − rSn = a − arn, and dividing by 1 − r finishes it. ∎

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, a(rn − 1)/(r − 1), keeping everything positive.

The last term alone is 3 × 29 = 1536, over half the total; geometric sums live in their final terms.

The infinite sum

When |r| < 1 the powers of r die away, the term arn in the sum formula vanishes as n grows, and the sum settles on a finite value,

S∞ = a1 − rIN THE FORMULAE BOOKLET

which is the opening lie's downfall. Endless additions can have a finite total, provided each addition is a fixed fraction of the last.

A bar of total length 16 filled by the segments 8, 4, 2, 1 and so on, each half the one before: the pieces never spill over the end, and the sum to infinity is the length of the box84218 + 4 + 2 + 1 + … lives inside a box of 16each term halves; the total never arrives, and never leaves
FIG. 18 + 4 + 2 + 1 + … packed into a box of length 16. The pieces crowd the end without crossing it: the sum to infinity is the box.

YOUR TURN

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, and |½| < 1, so the series converges.

S∞ = 8/(1 − ½) = 16.

The convergence sentence is a mark on its own; the formula only applies once |r| < 1 has been said.

Logs answer how long

Questions asking how many terms, or how many years of compound growth, put the unknown in an exponent, and the logarithms lesson takes over from there.

Two thousand pounds growing at 4 percent compound: the yearly values form a geometric sequence that first passes double the stake, 4000 pounds, in year 18£4000: money doubled4% compound growthyear 18
FIG. 2£2000 at 4% compound. The values are geometric with r = 1.04, and the first dot above £4000 is year 18, where the logarithm said it would be.

TRY IT UNSEEN

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: 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 and verify both neighbours, exactly as in the saving scheme, because the crossing itself is what the question is marking.

THE EXAM BIT

  • Identify a and r first, r from the ratio of consecutive terms, and state them before any formula.
  • Quote the sum formula and remember its subtraction proof; “prove the formula” is a standard opener.
  • Say |r| < 1 before using the sum to infinity; the condition is marked separately from the calculation.
  • How-many-terms questions end with a logarithm, a round-up, and a check of the two neighbouring values.
  • With r > 1, write the sum as a(rn − 1)/(r − 1) and keep the arithmetic positive.

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; both formulae then run on autopilot.

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 already carry nearly ninety per cent of the infinite total, which is how quickly a ratio of 0.8 fades.

Each term is r times the last; the sum telescopes when you subtract r times itself.

|r| < 1 buys convergence and a over 1 minus r; logs answer every how-long question.

CHECK YOUR PROGRESS

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  • Use the nth-term formula arn−1, and prove and use the finite sum formula.
  • Recognise convergence and use the sum to infinity when |r| < 1.
  • Solve how-many-terms questions with logarithms, in series and compound-growth models.

No animated video for this topic yet; these notes stand alone.