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AI HL · 12.1

Sequences & Series

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12.1.7

Infinite geometric series: the sum to infinity

12.1.7

Infinite geometric series: the sum to infinity

Target: I can say when an infinite geometric series has a sum, and explain why it has none when |r| is 1 or more. (+3 more)

By the end of this section

Key points

When does an infinite series have a sum?

  • In a geometric series \(u_1+u_1r+u_1r^2+\dots\) the ratio is \(r=\frac{u_2}{u_1}\). Check it with the next pair of terms too.
  • If \(-1<r<1\) (written \(|r|<1\)) the terms shrink towards 0 and the running total settles on one value. The series converges.
  • If \(|r|\ge 1\) the terms do not shrink. The total grows without limit, or swings back and forth, so there is no sum to infinity.
  • Picture: \(1+\frac12+\frac14+\dots\) has running totals 1, 1.5, 1.75, 1.875 and so on. They climb towards 2 and never pass it.

The sum to infinity

  • If \(|r|<1\): \(S_\infty=\frac{u_1}{1-r}\). This is in the formula booklet. Check \(|r|<1\) before you use it.
  • Find \(u_1\) and r first, write them down, then substitute. A common slip is to divide by r instead of by \(1-r\).
  • A negative r is allowed. The terms alternate in sign and \(S_\infty\) is smaller than \(u_1\).
  • When the terms are fractions, keep the answer as an exact fraction unless the question asks for a decimal.

Recurring decimals

  • A recurring decimal is an infinite geometric series. For \(0.\overline{45}\) the series is \(0.45+0.0045+0.000045+\dots\)
  • The first term is the repeating block written as a decimal. The ratio is \(\frac{1}{10^k}\), where k is the length of the block.
  • Then \(S_\infty\) is the fraction. Cancel it to lowest terms and check by dividing on the calculator.
  • If some digits come before the repeat, as in \(0.3\overline{18}\), write the non-repeating part as a fraction and add the series for the rest.

Two values of r from \(u_2\) and \(S_\infty\)

  • You know \(u_2=u_1r\), so \(u_1=\frac{u_2}{r}\). Put this into \(S_\infty=\frac{u_1}{1-r}\).
  • Multiply out to get a quadratic in r. Solve it with the calculator.
  • Test each root against \(|r|<1\). A root with \(|r|\ge 1\) is rejected. Often both roots pass, and they give two different series with the same sum.
  • For each r that passes, find \(u_1=\frac{u_2}{r}\) and check that \(\frac{u_1}{1-r}\) gives the sum you were told.

On the calculator

Solving the quadratic for r

Menu > Algebra > Polynomial Tools > Find Roots

  • Rearrange to \(ar^2+br+c=0\), with 0 on the right-hand side.
  • Enter the coefficients a, b, c in the polynomial tools and ask for the roots.
  • Read both roots. Keep only those with \(|r|<1\).

Screenshot and video to come

Two values of r by graph

  • Graph \(y=S\,x(1-x)\) on one side and \(y=u_2\) on the other, where S and \(u_2\) are the numbers in the question.
  • Find where the two graphs meet. The x-values are the values of r.
  • Use this to check the roots from the solver.

Screenshot and video to come

Checking a recurring decimal

  • Divide the numerator by the denominator of your fraction.
  • The decimal on the screen should start with the repeating digits you began with.

Screenshot and video to come

Explore: the sum to infinity

The dots are the running totals Sₙ of a geometric series. Choose u₁ and r, and say where the dots will settle, then press Show answer.

S₅
S₁₀
S₂₀
S∞ − S₂₀

Move a slider to change the series.

  1. Keep u₁ = 8 and move r from 0 up to 1. What happens to the line, and to how fast the dots reach it?
  2. Make r negative. How do the dots move, and why is S∞ less than u₁?
  3. Move r past 1, then past −1. What changes in the picture, and why is there no sum?
  4. Set r = 0.9. How many terms do you need before Sₙ is close to S∞? What does that tell you about r close to 1?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Does the series have a sum?

Find the common ratio of the series \(8+12+18+\dots\) Does it have a sum to infinity?

Working
  1. Ratio from two terms

    \(r=\frac{12}{8}=1.5\)

    Second term divided by first term.

  2. Check with the next pair

    \(\frac{18}{12}=1.5\)

    The same ratio, so the series is geometric.

  3. Is |r| < 1?

    \(|1.5|>1\)

    The terms grow. The running total grows without limit, so there is no sum to infinity.

Video walkthrough coming
Sum from fractions

Find the sum to infinity of \(\frac56+\frac{5}{18}+\frac{5}{54}+\dots\)

Working
  1. First term and ratio

    \(u_1=\frac56,\quad r=\frac{5/18}{5/6}=\frac13\)

    Divide the second term by the first.

  2. Check |r| < 1

    \(\left|\frac13\right|<1\)

    So the series converges.

  3. Substitute

    \(S_\infty=\frac{5/6}{1-\frac13}=\frac{5/6}{2/3}\)

    Write \(1-r\) as one fraction first.

  4. Answer

    \(S_\infty=\frac54=1.25\)

    Exact value \(\frac54\).

Video walkthrough coming
A recurring decimal as a fraction

Write \(0.\overline{45}\) as a fraction in lowest terms.

Working
  1. Write it as a series

    \(0.\overline{45}=0.45+0.0045+0.000045+\dots\)

    Two digits repeat, so each term is \(\frac{1}{100}\) of the one before.

  2. First term and ratio

    \(u_1=0.45=\frac{45}{100},\quad r=\frac{1}{100}\)

    |r| < 1, so the sum exists.

  3. Sum to infinity

    \(S_\infty=\frac{45/100}{1-\frac{1}{100}}=\frac{45/100}{99/100}=\frac{45}{99}\)

    The 100s cancel.

  4. Lowest terms

    \(0.\overline{45}=\frac{5}{11}\)

    Cancel by 9. From the calculator: \(5\div 11=0.4545\dots\), which matches.

Video walkthrough coming
Two values of r

A geometric series has second term 3 and sum to infinity 16. Find the two possible values of the common ratio.

Working
  1. First term in terms of r

    \(u_2=u_1r=3\Rightarrow u_1=\frac3r\)

    Use the second term to remove \(u_1\).

  2. Put into the sum

    \(\frac{3/r}{1-r}=16\Rightarrow 3=16r(1-r)\)

    Multiply both sides by \(r(1-r)\).

  3. Quadratic

    \(16r^2-16r+3=0\)

    Write it with 0 on the right-hand side.

  4. Solve

    From the calculator: \(r=0.25\) or \(r=0.75\)

    Polynomial tools, Find Roots.

  5. Test each

    \(|0.25|<1\) and \(|0.75|<1\). \(u_1=12\) or \(u_1=4\)

    Both are allowed. Check: \(\frac{12}{0.75}=16\) and \(\frac{4}{0.25}=16\).

Video walkthrough coming