AP Calculus BC glossary

Geometric Sequence

Also called: Geometric progression

A geometric sequence has a constant ratio r between consecutive terms, so the nth term is the first term times r to the power n minus 1. It converges when the absolute value of r is less than 1, and also when r equals 1, where every term is the same number.

an=a1r n−1a_n = a_1 r^{\,n-1}

Everything rides on rnr^n. For ∣r∣<1|r| < 1 the powers shrink and the sequence converges to 00. For r=1r = 1 every term equals a1a_1, so the sequence sits still, which counts as convergence. For r=−1r = -1 the terms flip between a1a_1 and −a1-a_1, and for ∣r∣>1|r| > 1 the terms grow without bound in size, running to infinity when r>1r > 1 and swinging through ever larger positive and negative values when r<−1r < -1.

lim⁡n→∞a1r n−1={0,∣r∣<1a1,r=1\lim_{n \to \infty} a_1 r^{\,n-1} = \begin{cases} 0, & |r| < 1 \\ a_1, & r = 1 \end{cases}

The corresponding series is stricter. ∑a1rn−1\sum a_1 r^{n-1} converges only for ∣r∣<1|r| < 1, to a1/(1−r)a_1/(1-r). At r=1r = 1 the sequence 5,5,5,…5, 5, 5, \ldots converges to 55 while the series 5+5+5+⋯5 + 5 + 5 + \cdots runs to infinity. That single value of rr is where the two conditions come apart.

The mistake

Importing the series condition ∣r∣<1|r| < 1 into a question about the sequence and declaring r=1r = 1 divergent. A constant sequence converges perfectly well. It is the sum of a constant sequence that does not.

Appears in: Unit 10: Infinite Sequences and Series (BC)