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1.11 Sum of infinite geometric sequencesIB Maths: Applications and Interpretation HL: Revision notes

Section 1

Geometric sequences and series

A geometric sequence has a constant ratio rr between terms: un=u1rn−1u_n=u_1r^{n-1}. The sum of the first nn terms is Sn=u1(1−rn)1−r,r≠1.S_n=\frac{u_1\left(1-r^{n}\right)}{1-r},\qquad r\ne1. Example: u1=12u_1=12, r=13r=\frac13 gives u4=12(13)3=49u_4=12\left(\frac13\right)^{3}=\frac49 and Sn=18(1−(13)n)S_n=18\left(1-\left(\frac13\right)^{n}\right). These formulae are in the formula booklet.

Key termsgeometric sequencecommon ratio

Section 2

When an infinite sum exists

If ∣r∣<1|r|<1 then rn→0r^{n}\to0 as n→∞n\to\infty, so the partial sums SnS_n settle on a limit. The series is convergent and its sum to infinity is the limit of SnS_n: S∞=u11−r,∣r∣<1.S_\infty=\frac{u_1}{1-r},\qquad|r|<1. If ∣r∣≥1|r|\ge1 the terms do not shrink, the series is divergent and there is no sum to infinity. For u1=12u_1=12 and r=13r=\frac13: S∞=1223=18S_\infty=\frac{12}{\frac23}=18. The partial sums S5=17.93…S_5=17.93\ldots are already very close.

Key termsconvergentdivergentsum to infinitylimit
Common mistake

Using the formula when ∣r∣≥1|r|\ge1. Always check ∣r∣<1|r|<1 before writing S∞S_\infty.

Common mistake

Writing 1+r1+r in the denominator. It is 1−r1-r.

Section 3

Recurring decimals

A recurring decimal is an infinite geometric series. For 0.272727…=0.27+0.0027+0.000027+…0.272727\ldots=0.27+0.0027+0.000027+\ldots the first term is u1=0.27u_1=0.27 and the ratio is r=0.01r=0.01, so S∞=0.271−0.01=0.270.99=2799=311.S_\infty=\frac{0.27}{1-0.01}=\frac{0.27}{0.99}=\frac{27}{99}=\frac{3}{11}. For 0.7777…0.7777\ldots: u1=0.7u_1=0.7, r=0.1r=0.1, so S∞=0.70.9=79S_\infty=\frac{0.7}{0.9}=\frac79. Group the repeating block to find u1u_1, and the ratio is 10−k10^{-k} for a block of kk digits.

Key termsrecurring decimal

Section 4

Modelling with infinite series

Many situations repeat a fixed percentage change. A ball dropped from 22 m that rises to 80%80\% of its previous height has rebound heights 1.6,1.28,…1.6,1.28,\ldots with r=0.8r=0.8. The total distance is 2+2×1.61−0.8=182+2\times\frac{1.6}{1-0.8}=18 m, because every rebound is travelled twice (up and down). For a daily drug dose of 200200 mg with 70%70\% remaining after each day, the amount just after the nnth dose is Sn=200(1−0.7n)0.3S_n=\frac{200\left(1-0.7^{n}\right)}{0.3}. It approaches 2000.3=667\frac{200}{0.3}=667 mg and never exceeds it. To find when a level is passed, solve Sn>kS_n>k with your GDC (table or logarithms).

Key termspartial sum
Exam tip

Write down u1u_1 and rr first, then decide whether you need SnS_n or S∞S_\infty.

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Exam questions on 1.11 Sum of infinite geometric sequences

  1. A geometric sequence has first term u1=12u_1=12 and common ratio r=13r=\frac13.
    Use your GDC to find the least value of nn for which the sum of the first nn terms is greater than 17.917.9.2 marks
  2. A ball is dropped from a height of 22 m onto a hard floor. After each bounce it rises to 80%80\% of the height from which it last fell.
    Use your GDC to find the number of the first bounce after which the ball rises to less than 0.10.1 m.2 marks
  3. The recurring decimal 0.272727…0.272727\ldots can be written as the sum of an infinite geometric series.
    Write down the first term and the common ratio of the series 0.27+0.0027+0.000027+…0.27+0.0027+0.000027+\ldots, and explain why its sum to infinity exists.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).