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10.2 Working with Geometric Series

Syllabus
2020
Topic
10.2
Level
—

Learning objectives

—LIM-7.A—Determine whether a series converges or diverges—Topic 10.2• LIM-7.A Determine whether a series converges or diverges.• LIM-7.A.3 A geometric series is a series with a constant ratio between successive terms.• LIM-7.A.4 If a is a real number and r is a real number such that |r| < 1, then the geometric series Σ(n=0 to ∞) arⁿ = a/(1 − r).• Enduring understanding LIM-7: Applying limits may allow us to determine the finite sum of infinitely many terms.

A Geometric Series Converges Only When Its Ratio Shrinks

A geometric series has the form a+ar+ar2+⋯a+ar+ar^2+\cdots, so dividing any term after the first by its preceding nonzero term gives the same common ratio rr. For a nonzero series, its terms shrink toward zero exactly when ∣r∣<1|r|<1.

\sum_{n=0}^{\infty}ar^n=\frac{a}{1-r}\quad\text{when }|r|<1;\qquad |r|\ge1\Rightarrow\text{divergence}

The finite partial sum through exponent NN is SN=a(1−rN+1)/(1−r)S_N=a(1-r^{N+1})/(1-r) for r≠1r\ne1. When ∣r∣<1|r|<1, rN+1→0r^{N+1}\to0, leaving a/(1−r)a/(1-r). If ∣r∣≥1|r|\ge1, that power does not approach zero, so the partial sums do not approach a finite value.

Example: 6−3+32−34+⋯6-3+\tfrac32-\tfrac34+\cdots is geometric because each term is the previous term multiplied by −1/2-1/2. Here a=6a=6 and r=−1/2r=-1/2. Since ∣−1/2∣<1|-1/2|<1, the series converges, and
S=61−(−1/2)=63/2=4.S=\frac{6}{1-(-1/2)}=\frac{6}{3/2}=4.
The alternating signs do not prevent convergence because the term magnitudes shrink.

Use the actual first term of the displayed series as aa. For example, ∑n=1∞3(1/2)n\sum_{n=1}^{\infty}3(1/2)^n begins with 3/23/2, not 33; either reindex from zero or use a=3/2a=3/2. Also, r=−1r=-1 alternates but does not shrink, so it diverges.