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AI HL · 12.1

Sequences & Series

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12.1.8

Infinite series: the range of x, r and the first term

12.1.8

Infinite series: the range of x, r and the first term

Target: I can find the sum to infinity from \(u_1\) and r, including when r is negative. (+5 more)

By the end of this section

Key points

Sum to infinity: the checks

  • Find r first, as any term divided by the term before. Then check \(|r|<1\). Only then use \(S_\infty=\frac{u_1}{1-r}\).
  • A negative ratio is fine. Keep the sign: \(1-r\) with \(r=-\frac13\) is \(1+\frac13\).
  • In sigma notation the first term is what you get when you put in the lowest value of k. For \(\sum_{k=1}^{\infty}12\left(\frac13\right)^k\) the first term is 4, not 12.

The range of x for convergence

  • When r contains x, the series converges only for some x. Solve \(-1<r<1\), which is the same as \(|r|<1\).
  • Add or subtract first. When you multiply or divide an inequality by a negative number, reverse both inequality signs.
  • If r contains \(x^2\), such as \(r=3x^2\), take square roots at the end: \(x^2<\frac13\) gives \(-\frac{1}{\sqrt3}<x<\frac1{\sqrt3}\).
  • If r is a fraction with x underneath, such as \(r=\frac2x\), then \(|r|<1\) means \(|x|>2\), so \(x<-2\) or \(x>2\).
  • Write the range as an inequality, for example \(2<x<3\). Outside the range there is no sum.

\(S_\infty\) as an expression in x

  • Put \(u_1\) and the ratio, which contains x, into \(\frac{u_1}{1-r}\). Simplify the fraction.
  • The expression is only true for x inside the range. Check that x is inside before you put a number in.
  • To find \(S_\infty\) at a given x, you can also work out r at that x and use the formula with numbers.

Finding the unknowns

  • r from \(u_1\) and \(S_\infty\): \(1-r=\frac{u_1}{S_\infty}\). \(u_1\) from \(S_\infty\) and r: \(u_1=S_\infty(1-r)\).
  • From \(u_2\) and \(S_\infty\): use \(u_1=\frac{u_2}{r}\), get \(u_2=S_\infty\,r(1-r)\), and solve the quadratic. Reject any r with \(|r|\ge 1\). A negative \(u_2\) usually leaves only one r.
  • From \(S_3\) and \(S_\infty\): \(S_3=S_\infty(1-r^3)\), so \(r^3=1-\frac{S_3}{S_\infty}\). Find r, then \(u_1=S_\infty(1-r)\).
  • With an unknown x: write the ratio and first term in terms of x, put them in \(\frac{u_1}{1-r}\), set equal to the sum, and solve for x. Check x is in the range.

On the calculator

Solving an equation for x or r

  • Write the equation exactly as it is, with the fraction on the left and the sum on the right.
  • Use the calculator's solve command for the unknown.
  • Read every solution, then test each one against \(|r|<1\).

Screenshot and video to come

Two values of r from a quadratic

Menu > Algebra > Polynomial Tools > Find Roots

  • Write \(S_\infty\,r(1-r)=u_2\) as \(ar^2+br+c=0\).
  • Find the roots in the polynomial tools.
  • Reject any root with \(|r|\ge 1\). Find \(u_1=\frac{u_2}{r}\) for each root that is left.

Screenshot and video to come

Checking a range by graph

  • Graph \(y=|r|\), with r written as an expression in x, and the line \(y=1\).
  • The series converges where the curve is below the line. Read the crossing points as the ends of the range.

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.
A linear ratio

The series \((2x-5)+(2x-5)^2+(2x-5)^3+\dots\) converges. Find the range of values of x.

Working
  1. The ratio

    \(r=2x-5\)

    Each term is the one before times \((2x-5)\).

  2. Convergence

    \(-1<2x-5<1\)

    \(|r|<1\).

  3. Add 5

    \(4<2x<6\)

    Do the same to all three parts.

  4. Divide by 2

    \(2<x<3\)

    Dividing by a positive number keeps the signs.

Video walkthrough coming
A negative coefficient

Find the range of x for which \(3-\frac{6x}{5}+\frac{12x^2}{25}-\dots\) converges.

Working
  1. The ratio

    \(r=\frac{-6x/5}{3}=-\frac{2x}{5}\)

    Second term divided by first. Check: \(\frac{12x^2/25}{-6x/5}=-\frac{2x}{5}\).

  2. Convergence

    \(-1<-\frac{2x}{5}<1\)

    \(|r|<1\).

  3. Multiply by \(-\frac52\)

    \(\frac52>x>-\frac52\)

    A negative multiplier reverses both signs.

  4. Tidy

    \(-2.5<x<2.5\)

    Write the smaller number on the left.

Video walkthrough coming
A ratio with \(x^2\)

Find the range of x for which \(2+6x^2+18x^4+\dots\) converges.

Working
  1. The ratio

    \(r=\frac{6x^2}{2}=3x^2\)

    Check: \(\frac{18x^4}{6x^2}=3x^2\).

  2. Convergence

    \(|3x^2|<1\Rightarrow 3x^2<1\)

    \(3x^2\) is never negative, so the left side adds nothing.

  3. Solve

    \(x^2<\frac13\Rightarrow -\frac{1}{\sqrt3}<x<\frac{1}{\sqrt3}\)

    A square root gives two limits.

  4. Decimals

    \(-0.577<x<0.577\)

    3 significant figures.

Video walkthrough coming
S∞ in terms of x

A series is \(8+8(1-2x)+8(1-2x)^2+\dots\) (a) Find the range of x for which it converges. (b) Find \(S_\infty\) as an expression in x. (c) Find \(S_\infty\) when \(x=0.4\).

Working
  1. (a) Range

    \(-1<1-2x<1\Rightarrow 0<x<1\)

    Subtract 1: \(-2<-2x<0\). Divide by \(-2\) and reverse: \(1>x>0\).

  2. (b) Expression

    \(S_\infty=\frac{8}{1-(1-2x)}=\frac{8}{2x}=\frac4x\)

    Simplify the bottom: \(1-(1-2x)=2x\).

  3. (c) At 0.4

    \(x=0.4\) is in the range, so \(S_\infty=\frac{4}{0.4}=10\)

    Check: \(r=0.2\) and \(\frac{8}{0.8}=10\).

Video walkthrough coming
Finding r, then \(u_1\)

(a) A series has \(u_1=12\) and \(S_\infty=20\). Find r. (b) A different series has \(S_\infty=15\) and \(r=-\frac13\). Find \(u_1\) and write down the first three terms.

Working
  1. (a) Rearrange

    \(20=\frac{12}{1-r}\Rightarrow 1-r=\frac{12}{20}\)

    Swap \(1-r\) and 20.

  2. (a) r

    \(r=1-0.6=0.4\)

    \(|0.4|<1\), so it is allowed.

  3. (b) \(u_1\)

    \(u_1=15\left(1-\left(-\frac13\right)\right)=15\times\frac43=20\)

    Multiply \(S_\infty\) by \(1-r\).

  4. (b) Terms

    \(20,\ -\frac{20}{3},\ \frac{20}{9}\)

    Multiply by \(-\frac13\) each time.

Video walkthrough coming
Two values of \(u_1\)

A geometric series has second term 5 and sum to infinity 36. (a) Find the two possible values of r. (b) Hence find the two possible values of \(u_1\).

Working
  1. Quadratic

    \(u_1=\frac5r\), so \(\frac{5/r}{1-r}=36\Rightarrow 36r^2-36r+5=0\)

    Multiply out and collect on the left.

  2. (a) Solve

    From the calculator: \(r=\frac16\) or \(r=\frac56\)

    Both have \(|r|<1\).

  3. (b) \(u_1=\frac5r\)

    \(r=\frac16\Rightarrow u_1=30\); \(r=\frac56\Rightarrow u_1=6\)

    Smaller r, larger first term.

  4. Check

    \(\frac{30}{5/6}=36\) and \(\frac{6}{1/6}=36\)

    Both series have sum 36.

Video walkthrough coming
Sigma notation

A series is \(\sum_{k=0}^{\infty}5\left(-\frac23\right)^k\). (a) Write down the first three terms. (b) Find the sum to infinity.

Working
  1. (a) Put in k = 0, 1, 2

    \(5,\ 5\left(-\frac23\right)=-\frac{10}{3},\ 5\left(-\frac23\right)^2=\frac{20}{9}\)

    The lowest value of k gives the first term.

  2. First term and ratio

    \(u_1=5,\quad r=-\frac23\)

    The number being raised to the power k is the ratio.

  3. (b) Sum

    \(S_\infty=\frac{5}{1-\left(-\frac23\right)}=\frac{5}{5/3}=3\)

    \(\left|-\frac23\right|<1\), so the sum exists.

Video walkthrough coming
Using \(S_3\) and \(S_\infty\)

A geometric series has sum to infinity 24. The sum of its first three terms is 21. Find the first term.

Working
  1. Link the two sums

    \(S_3=\frac{u_1(1-r^3)}{1-r}=S_\infty(1-r^3)\)

    Both sums have the same \(\frac{u_1}{1-r}\).

  2. Find \(r^3\)

    \(21=24(1-r^3)\Rightarrow r^3=\frac18\)

    Divide by 24 and rearrange.

  3. Find r

    \(r=\frac12\)

    Cube root.

  4. Find \(u_1\)

    \(u_1=24\left(1-\frac12\right)=12\)

    Check: \(12+6+3=21\).

Video walkthrough coming