Unit 10 · Topic 10.2

Geometric Series

A geometric series ∑ ar^n has a constant ratio r between terms.

Practice 1: Implementing Mathematical ProcessesPractice 3: Justification

Part 1 · Hook

Why this matters

Cut a square in half, then cut one half in half, and keep going: the pieces 12+14+18+⋯\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots fill the whole square, exactly 1. Each piece is the one before times the same number. Series built this way, geometric series, are the one family where you can always write the sum, and they return in every later topic of the unit.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. A series has partial sums Sn=2−1nS_n = 2 - \frac{1}{n}. Its sum is

  1. 22
  2. 11
  3. 00
  4. It has no sum
Show the answer

The sum is limn→∞Sn=2\lim_{n \to \infty} S_n = 2.

  • Correct: 22:
  • 11:
  • 00:
  • It has no sum:

2. limn→∞(34)n=\lim_{n \to \infty} \left(\frac{3}{4}\right)^n =

  1. 00
  2. 11
  3. 34\frac{3}{4}
  4. ∞\infty
Show the answer

A number between −1-1 and 11 raised to larger and larger powers shrinks to 0.

  • Correct: 00:
  • 11:
  • 34\frac{3}{4}:
  • ∞\infty:

Part 4 · See it

See it first

Partial sums 1 + r + ... + r to the n (the first n + 1 terms) for n = 0 to 10, one dot each, for three ratios. Left, r = 1/2: the partial sums 1, 1.5, 1.75, ... rise toward the dashed line at 2, the sum 1/(1 − 1/2). Middle, r = −1/2: the partial sums 1, 0.5, 0.75, 0.625, ... zigzag in toward 2/3, the sum 1/(1 + 1/2). Right, r = 5/4: the partial sums grow without bound, so the series diverges.
Partial sums for r=12r = \frac{1}{2} and r=−12r = -\frac{1}{2} settle at 11−r\frac{1}{1 - r}; for r=54r = \frac{5}{4} they grow without bound. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Each term is the previous one times rrSnS_n=a= a+ar+ ar+⋯+ \cdots+arn−1+ ar^{n-1}=a(1−rn)1−r= \frac{a(1 - r^n)}{1 - r}
  2. If |r|<1|r| < 1, then rn→0r^n \to 0Sn→a1−rS_n \to \frac{a}{1 - r}: the series converges
  3. If |r|≥1|r| \ge 1, the terms do not shrink to 0the partial sums have no limit: the series diverges
  4. The sum starts at a later indexaa is the first term actually in the sum, not the value at n=0n = 0

Part 6 · Key ideas

Key ideas

  • ∑n=0∞arn\sum_{n=0}^{\infty} ar^n=a1−r= \frac{a}{1 - r} when |r|<1|r| < 1; it diverges when |r|≥1|r| \ge 1 (and a≠0a \ne 0).
  • In the formula, aa is the first term of the sum as written: ∑n=1∞5(14)n\sum_{n=1}^{\infty} 5\left(\frac{1}{4}\right)^n has a=54a = \frac{5}{4}, not 5.
  • Find rr by dividing a term by the one before it; rewrite 2n3n+1\frac{2^n}{3^{n+1}} as 13(23)n\frac{1}{3}\left(\frac{2}{3}\right)^n to see aa and rr.
  • With xx in the ratio, the series converges for the xx where |r|<1|r| < 1, and its sum is a function of xx.

Part 7 · Misconception

A common mistake

The wrong idea: ∑n=2∞3(12)n\sum_{n=2}^{\infty} 3\left(\frac{1}{2}\right)^n=31−1/2= \frac{3}{1 - 1/2}=6= 6.

What actually happens: The first term in the sum is at n=2n = 2: 3⋅14=343 \cdot \frac{1}{4} = \frac{3}{4}. The sum is 3/41/2=32\frac{3/4}{1/2} = \frac{3}{2}.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

The first terms of a geometric series

The terms ana_n of a geometric series ∑n=0∞an\sum_{n=0}^{\infty} a_n, starting at n=0n = 0
nn00112233
ana_n1212−8-8163\frac{16}{3}−329-\frac{32}{9}

1. Find the common ratio rr.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Divide any term by the one before it: −812=−23\frac{-8}{12} = -\frac{2}{3}, and 16/3−8=−23\frac{16/3}{-8} = -\frac{2}{3}.

  • Answer: −23\frac{-2}{3} ≈ -0.667

2. Find ∑n=0∞an\sum_{n=0}^{\infty} a_n.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

a1−r\frac{a}{1 - r}=121−(−23)= \frac{12}{1 - \left(-\frac{2}{3}\right)}=125/3= \frac{12}{5/3}=365= \frac{36}{5}.

  • Answer: 365\frac{36}{5} ≈ 7.200

3. Find ∑n=2∞an\sum_{n=2}^{\infty} a_n.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Starting at n=2n = 2, the first term is a2=163a_2 = \frac{16}{3} and the ratio is still −23-\frac{2}{3}: 16/35/3=165\frac{16/3}{5/3} = \frac{16}{5}. Check: 365−12+8=165\frac{36}{5} - 12 + 8 = \frac{16}{5}.

  • Answer: 165\frac{16}{5} ≈ 3.200

Table

A geometric series with xx in the ratio

The first terms of ∑n=0∞(x−13)n\sum_{n=0}^{\infty} \left(\frac{x - 1}{3}\right)^n
nn00112233
Term11x−13\frac{x - 1}{3}(x−1)29\frac{(x - 1)^2}{9}(x−1)327\frac{(x - 1)^3}{27}

4. Give all xx for which the series converges, as an interval.

Type the interval, in interval notation such as [1, 5) or as an inequality such as 1 ≤ x < 5.

Show the answer

The series converges when |x−13|<1\left|\frac{x - 1}{3}\right| < 1, that is |x−1|<3|x - 1| < 3, or −2<x<4-2 < x < 4. At both ends |r|=1|r| = 1, and the series diverges, so the interval is (−2,4)(-2, 4).

  • Answer: (−2,4)(-2, 4)

5. For −2<x<4-2 < x < 4, write the sum of the series as an expression in xx.

Type an expression for ∑n=0∞(x−13)n=\sum_{n=0}^{\infty} \left(\frac{x - 1}{3}\right)^n =.

Show the answer

11−x−13\frac{1}{1 - \frac{x - 1}{3}}=33−(x−1)= \frac{3}{3 - (x - 1)}=34−x= \frac{3}{4 - x}.

  • Answer: ∑n=0∞(x−13)n=\sum_{n=0}^{\infty} \left(\frac{x - 1}{3}\right)^n = 34−x\frac{3}{4 - x}

6. Find ∑n=1∞5(14)n\sum_{n=1}^{\infty} 5\left(\frac{1}{4}\right)^n.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

The first term is at n=1n = 1: a=5⋅14=54a = 5 \cdot \frac{1}{4} = \frac{5}{4}. Sum: 5/41−1/4\frac{5/4}{1 - 1/4}=5/43/4= \frac{5/4}{3/4}=53= \frac{5}{3}.

  • Answer: 53\frac{5}{3} ≈ 1.667

Part 9 · Summary

Summary

A geometric series ∑arn\sum ar^n has a constant ratio rr between terms. It converges exactly when |r|<1|r| < 1, to first term1−r\frac{\text{first term}}{1 - r}, where the first term is the first one actually in the sum. If |r|≥1|r| \ge 1 it diverges.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Question set for this topic