Geometric Series (College Board AP® Calculus BC): Study Guide

Roger B

Written by: Roger B

Reviewed by: Mark Curtis

Updated on

Geometric series

What is a geometric series?

  • A geometric series is a series of the form sum from n equals 0 to infinity of a r to the power of n equals a plus a r plus a r squared plus...

    • where a not equal to 0 is a real number constant

      • a is also the first term of the series

    • and r is a real number known as the common ratio

    • Note that the series starts with bold italic n bold equals bold 0

  • Note that there is a constant ratio between one term and the next

    • For example sum from n equals 0 to infinity of 9 times open parentheses 1 third close parentheses to the power of n equals 9 plus 3 plus 1 plus 1 third plus...

      • 3 equals 1 third times 9, space 1 equals 1 third times 3, space 1 third equals 1 third times 1, etc.

      • Each term is the preceding term multiplied by the common ratio 1 third

When does a geometric series converge?

  • The sum of the first bold italic n bold plus bold 1 terms of a geometric series is given by

    • s subscript n equals fraction numerator a open parentheses 1 minus r to the power of n plus 1 end exponent close parentheses over denominator 1 minus r end fraction

      • I.e. space s subscript 0 equals fraction numerator a open parentheses 1 minus r to the power of 0 plus 1 end exponent close parentheses over denominator 1 minus r end fraction equals fraction numerator a open parentheses 1 minus r to the power of 1 close parentheses over denominator 1 minus r end fraction equals a

      • space s subscript 1 equals fraction numerator a open parentheses 1 minus r to the power of 1 plus 1 end exponent close parentheses over denominator 1 minus r end fraction equals fraction numerator a open parentheses 1 minus r squared close parentheses over denominator 1 minus r end fraction equals fraction numerator a open parentheses 1 minus r close parentheses open parentheses 1 plus r close parentheses over denominator 1 minus r end fraction equals a open parentheses 1 plus r close parentheses equals a plus a r

      • etc.

    • If r equals 1, then the series is a plus a plus a plus... so s subscript n equals a open parentheses n plus 1 close parentheses

  • If open vertical bar r close vertical bar less than 1 then the geometric series sum from n equals 0 to infinity of a r to the power of n converges

    • and sum from n equals 0 to infinity of a r to the power of n equals fraction numerator a over denominator 1 minus r end fraction

      • This comes from sum from n equals 0 to infinity of a r to the power of n equals limit as n rightwards arrow infinity of s subscript n equals limit as n rightwards arrow infinity of open parentheses fraction numerator a open parentheses 1 minus r to the power of n plus 1 end exponent close parentheses over denominator 1 minus r end fraction close parentheses equals fraction numerator a open parentheses 1 minus 0 close parentheses over denominator 1 minus r end fraction equals fraction numerator a over denominator 1 minus r end fraction

  • If open vertical bar r close vertical bar greater or equal than 1 then the geometric series diverges

    • For example if r equals 1, then the series is sum from n equals 0 to infinity of a times 1 to the power of n equals a plus a plus a plus a plus...

      • The sequence of partial sums is a comma space 2 a comma space 3 a comma space 4 a comma..., which diverges to plus-or-minus infinity (depending on whether a is positive or negative)

      • So the series diverges

    • Or if a equals 5 and r equals 2, then the series is sum from n equals 0 to infinity of 5 times 2 to the power of n equals 5 plus 10 plus 20 plus 40 plus...

      • Each term added on is bigger than the one before

      • The sequence of partial sums is 5 comma space 15 comma space 35 comma space 75 comma..., which diverges to plus infinity

      • So the series diverges

Worked Example

Consider the infinite series sum from n equals 1 to infinity of 1 over 2 to the power of n. Show that the series converges and determine the value of the sum.

Rewrite the series so that it has the standard form of a geometric series

table row cell sum from n equals 1 to infinity of 1 over 2 to the power of n end cell equals cell 1 half plus 1 fourth plus 1 over 8 plus... end cell row blank equals cell 1 half times open parentheses 1 half close parentheses to the power of 0 plus 1 half times open parentheses 1 half close parentheses to the power of 1 plus 1 half times open parentheses 1 half close parentheses squared plus.. end cell row blank equals cell sum from n equals 0 to infinity of 1 half times open parentheses 1 half close parentheses to the power of n end cell end table

Note that we've changed the sum to start with n equals 0

That is a geometric series with a equals 1 half and r equals 1 half

Show that it satisfies the convergence condition

That is a geometric series with open vertical bar r close vertical bar equals open vertical bar 1 half close vertical bar equals 1 half less than 1 so it converges

Use the formula sum from n equals 0 to infinity of a r to the power of n equals fraction numerator a over denominator 1 minus r end fraction

So sum from n equals 1 to infinity of 1 over 2 to the power of n equals sum from n equals 0 to infinity of 1 half times open parentheses 1 half close parentheses to the power of n equals fraction numerator 1 half over denominator 1 minus 1 half end fraction equals fraction numerator 1 half over denominator 1 half end fraction equals 1

sum from n equals 1 to infinity of 1 over 2 to the power of n equals 1

Worked Example

Consider the number 0.6 with dot on top 2 1 with dot on top equals 0.621621621621....

(a) Write the number in the form of a geometric series.

Use the fact that 0.6 with dot on top 2 1 with dot on top equals 0.621 plus 0.000621 plus 0.000000621 plus...

table row cell 0.6 with dot on top 2 1 with dot on top end cell equals cell 621 over 1000 plus 621 over 1000000 plus 621 over 1000000000 plus... end cell row blank equals cell 621 over 1000 plus 621 over open parentheses 1000 close parentheses squared plus 621 over open parentheses 1000 close parentheses cubed plus... end cell row blank equals cell 621 over 1000 times open parentheses 1 over 1000 close parentheses to the power of 0 plus 621 over 1000 times open parentheses 1 over 1000 close parentheses to the power of 1 plus 621 over 1000 times open parentheses 1 over 1000 close parentheses squared plus... end cell row blank equals cell sum from n equals 0 to infinity of 621 over 1000 times open parentheses 1 over 1000 close parentheses to the power of n end cell end table

table row cell 0.6 with dot on top 2 1 with dot on top end cell equals cell sum from n equals 0 to infinity of 621 over 1000 times open parentheses 1 over 1000 close parentheses to the power of n end cell end table

(b) Use the result from part (a) to find the value of 0.6 with dot on top 2 1 with dot on top as a fraction in lowest terms.

table row blank blank cell sum from n equals 0 to infinity of 621 over 1000 times open parentheses 1 over 1000 close parentheses to the power of n end cell end table is a geometric series with a equals 621 over 1000 and r equals 1 over 1000

First show that the series converges

For that geometric series, open vertical bar r close vertical bar equals open vertical bar 1 over 1000 close vertical bar equals 1 over 1000 less than 1 so it converges

Now you can use the formula sum from n equals 0 to infinity of a r to the power of n equals fraction numerator a over denominator 1 minus r end fraction

Therefore

table row cell sum from n equals 0 to infinity of 621 over 1000 times open parentheses 1 over 1000 close parentheses to the power of n end cell equals cell fraction numerator bevelled 621 over 1000 over denominator 1 minus bevelled 1 over 1000 end fraction end cell row blank equals cell fraction numerator bevelled 621 over 1000 over denominator bevelled 999 over 1000 end fraction end cell row blank equals cell 621 over 999 end cell row blank equals cell 69 over 111 end cell row blank equals cell 23 over 37 end cell end table

table row cell 0.6 with dot on top 2 1 with dot on top end cell equals cell 23 over 37 end cell end table

Alternative forms of a geometric series

How else can a geometric series be written?

  • You may sometimes see geometric series written with the sum starting at n equals 1

  • One example is sum from n equals 1 to infinity of a r to the power of n minus 1 end exponent equals a plus a r plus a r squared plus...

    • In this form most things remain the same

      • a is still the first term of the series

      • r is still the common ratio

      • The infinite sum still converges to fraction numerator a over denominator 1 minus r end fraction if open vertical bar r close vertical bar less than 1

    • But the formula for s subscript n is slightly different

      • s subscript n equals fraction numerator a open parentheses 1 minus r to the power of n close parentheses over denominator 1 minus r end fraction gives the sum of the first bold italic n terms of the series

  • Another example is sum from n equals 1 to infinity of a r to the power of n equals a r plus a r squared plus a r cubed plus...

    • In this form some things remain the same

      • r is still the common ratio

      • The infinite sum still only converges if open vertical bar r close vertical bar less than 1

    • But some other things change

      • a r is the first term of the series

      • The infinite sum, if it converges, converges to fraction numerator a over denominator 1 minus r end fraction minus a

      • s subscript n equals fraction numerator a open parentheses 1 minus r to the power of n close parentheses over denominator 1 minus r end fraction minus a space gives the sum of the first bold italic n terms of the series

  • You can work with these forms if you prefer

    • unless an exam question specifically indicates otherwise

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Roger B

Author: Roger B

Expertise: Maths

Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

Mark Curtis

Author: Mark Curtis

Expertise: Maths

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.