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CS221 Geometric Series

Contents

Introduction

A geometric series a + ar + ar^2 + ar^3 + \dots is a series in which there is a constant ratio r between consecutive terms. You can find the ratio by dividing a term by the one before it.

Example

  1. 1+2+4+8+16+32+\dots is a geometric series. The first term is a= 1 and the constant ratio, found by dividing consecutive terms is r =\frac{2}{1}=\frac{4}{2}=\frac{8}{4}=\frac{16}{8}=\frac{32}{16}=2. The nth term is given by 2^{n-1} for n\ge 1.
  2. 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \dots is a geometric series with a= 1 and r =\frac{1}{2} where the nth term is given by \frac{1}{2^{n-1}}.
  3. 2 + \frac{4}{3} + \frac{8}{9} + \frac{16}{27} + \frac{32}{81} + \dots is a geometric series with a= 2 and r =\frac{2}{3} where the nth term is given by \frac{2^n}{3^{n-1}}.
  4. 1 + x + x^2 + x^3 + x^4 + \dots is a geometric series with a= 1 and r =x where the nth term is given by x^{n-1}.
  5. 1 + x^2 + x^4 + x^6 + x^9 + \dots is a geometric series with a= 1 and r =x^2 where the nth term is given by x^{2(n-1)}.

Exercises

For each of the following geometric series, find a and r.

  1. 2 + 10 + 50 + 250 + 1250 + \dots
  2. 2 + \frac{1}{5} + \frac{1}{50} + \frac{1}{500} + \frac{1}{5000} + \dots
  3. 1 + x^4 + x^8 + x^{12} + x^{16} + \dots
  4. 1 + x^6 + x^{12} + x^{18} + x^{24} + \dots

Answers

For each of the following geometric series, find a and r.

  1. 2 + 10 + 50 + 250 + 1250 + \dots , a=2, r=5
  2. 2 + \frac{1}{5} + \frac{1}{50} + \frac{1}{500} + \frac{1}{5000} + \dots , a=2, r= \frac{1}{10}
  3. 1 + x^4 + x^8 + x^{12} + x^{16} + \dots, a=1, r=x^4
  4. 1 + x^6 + x^{12} + x^{18} + x^{24} + \dots, a=1, r=x^6

Sum of First n Terms

The sum of the first n terms of a geometric progression, where r \ne 1 is

a + ar + ar^2 + ar^3 + \dots + ar^{n-1} = \frac{a(1-r^n)}{1-r}.

Example

  1. For 1+2+4+8+16++32+ 64 + 128 we have a=1, r=2, n=8, so 1+2+4+8+16+64 + 128 = \frac{1(1-2^8)}{1-2} = \frac{1-256}{-1}= \frac{-255}{-1}= 255.
  2. 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} = \frac{1(1-\frac{1}{2}^6)}{1-\frac{1}{2}}= \frac{1(1-\frac{1}{64})}{\frac{1}{2}}= \frac{\frac{63}{64})}{\frac{1}{2}}=\frac{63}{32} since a=1, r=\frac{1}{2} and n=6
  3. 1 + x + x^2 + x^3 + x^4 + x^5 +x^6 =\frac{(1-x^7)}{1-x} since a=1, r=x and n=7
  4. 1 + x^2 + x^4 + x^6 + x^8 =\frac{1(1-(x^2)^5)}{1-x^2}=\frac{1-(x^{2\cdot5})}{1-x^2}= \frac{1-x^{10}}{1-x^2} as a= 1 and r =x^2 and n=5.

Exercises

Use the formula to find the following sums.

  1. 2 + 10 + 50 + 250 + 1250
  2. 2 + \frac{1}{5} + \frac{1}{50} + \frac{1}{500} + \frac{1}{5000}
  3. 1 + x
  4. 1 + x^4 + x^8 + x^{12} + x^{16}+x^{20}+x^{24}
  5. 1 + x^6 + x^{12} + x^{18} + x^{24} + x^{30}

Answers

  1. 2 + 10 + 50 + 250 + 1250 = \frac{2(1-5^5}{1-5}= \frac{2(1-3125)}{-4}=\frac{2(-3124)}{-4} =1562
  2. 2 + \frac{1}{5} + \frac{1}{50} + \frac{1}{500} + \frac{1}{5000}=\frac{2(1-(\frac{1}{10})^5)}{1-\frac{1}{10}}=2(-\frac{9999}{100000}){-\frac{9}{10}}=\frac{2222}{10000} (or \frac{1111}{5000} or 0.2222)
  3. 1 + x = \frac{1(1-x^2)}{1-x}=\frac{1-x^2}{1-x}
  4. 1 + x^4 + x^8 + x^{12} + x^{16}+x^{20}+x^{24} = \frac{1(1-(x^4)^7)}{1-x}=\frac{1-x^{28}}{1-x}
  5. 1 + x^6 + x^{12} + x^{18} + x^{24} + x^{30}= \frac{1(1-(x^6)^6)}{1-x}=\frac{1-x^{36}}{1-x}

Sum of Infinite Geometric Series

For -1 < r < 1 and a=1, r^n tends to zero and so 1-r^n \rightarrow 1. This means that

1 + r + r^2 + r^3 + r^4 + \dots = \frac{1}{1-r} for -1 < r < 1.

Example

  1. 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + \dots = \frac{1}{1-\frac{1}{3}}= \frac{1}{\frac{2}{3}}= \frac{3}{2}=1.5
  2. For -1<x<1, 1 + x + x^2 + x^3 + x^4 + x^5 +\dots =\frac{1}{1-x}
  3. 1 + x^2 + x^4 + x^6 + x^8 =\frac{1}{1-x^2} for -1 <x<1 (since the condition for r, -1<x^2<1 is satisfied exactly when -1 <x<1.)
  4. \frac{1}{1-x^3} = 1 + x^3 + (x^3)^2 + (x^3)^3 + (x^3)^4 + \dots = 1 + x^3 + x^6 + x^9 + x^12 + \dots for -1 <x<1

Exercises

A. Use the formula to find the following sums.

  1. 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}+\dots
  2. 1 + x^4 + x^8 + x^{12} + x^{16}+x^{20}+x^{24} for -1<x<1
  3. 1 + x^6 + x^{12} + x^{18} + x^{24} + x^{30} for -1<x<1

B. What are the geometric series for each of the following, where -1<x<1?

  1. \frac{1}{1-x}
  2. \frac{1}{1-x^2}
  3. \frac{1}{1-x^4}

Answers

A.

  1. 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}+\dots =\frac{1}{1-\frac{1}{2}} =\frac{1}{\frac{1}{2}} = 2
  2. 1 + x^4 + x^8 + x^{12} + x^{16}+x^{20}+x^{24} = \frac{1}{1-x^4} for -1<x<1
  3. 1 + x^6 + x^{12} + x^{18} + x^{24} + x^{30}\frac{1}{1-x^6} for -1<x<1

B.

  1. \frac{1}{1-x}= 1 + x + x^2 + x^3 + \ldots
  2. \frac{1}{1-x^2}-1 <x<1$=1 + x^2 + (x^2)^2 + (x^2)^3 + (x^2)^4 + \dots = 1 + x^2 + x^4 + x^6 + x^8 + \dots for -1<x<1
  3. \frac{1}{1-x^4}=1 + x^4 + (x^4)^2 + (x^4)^3 + (x^4)^4 + \dots = 1 + x^4 + x^8 + x^{12} + x^{16} + \dots for -1<x<1