A variable geometric ratio

The geometric series is given by 3x−24x+(3x−2)24x+(3x−2)34x+⋯\frac{3x-2}{4x}+\frac{(3x-2)^2}{4x}+\frac{(3x-2)^3}{4x}+\cdots where x≠0x\ne0. Explain why the sum to infinity exists for 13<x<1\frac13<x<1, and find its value in terms of xx. marks

[2020/DHS/Prelim/I/1]

Solution

Convergence and sum

Sum to infinity

3x−24x+(3x−2)24x+(3x−2)34x+⋯\frac{3x-2}{4x}+\frac{(3x-2)^2}{4x}+\frac{(3x-2)^3}{4x}+\cdots with x≠0x\ne0. Explain why the sum to infinity exists for 13<x<1\frac13<x<1 and find it in terms of xx.

Explore this idea

Hint

  1. Identify the common ratio rr.

  2. The sum to infinity exists iff ∣r∣<1|r|<1.

  3. Use S∞=a1−rS_\infty=\frac a{1-r}.

Answer · Convergence and sum

Converges for 1/3<x<11/3<x<1; sum (3x−2)/[12x(1−x)](3x-2)/[12x(1-x)].

Explore the idea

Terms and running total for x = 0.8

Drag the slider to change n and compare the plotted values.

Indexn = 5
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term uₙ at n = 5
0.0032
sum Sₙ at n = 5
0.2062
Key idea

With ratio r=0.4r=0.4 the terms shrink geometrically and the partial sums level off at S∞=a1−rS_\infty=\frac{a}{1-r}.

Ideas to try
Slide to a large nn: the running total settles at S∞=0.412(0.8)(0.2)≈0.208S_\infty=\frac{0.4}{12(0.8)(0.2)}\approx0.208.
Takeaway

A multiplier is safer than a quotient

To identify a GP, ask what multiplies each term to produce the next. A quotient of consecutive terms is unavailable when both terms are zero.