For more information, please see full course syllabus of Calculus BC

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For more information, please see full course syllabus of Calculus BC

For more information, please see full course syllabus of Calculus BC

### Geometric Series

- Series is the sum of a all terms in a sequence
- Geometric series is a series with a component raised to a higher power with each iteration
- If the component, , then higher power leads to divergence
- If the component, , then smaller fractions to higher powers leads to convergence

### Geometric Series

Does ∑

_{n = 1}^{∞}7(2)^{n − 1}converge or diverge?Determine rr = 2|2| > 1, thus the series diverges

Does ∑

_{n = 1}^{∞}[(3^{n})/3] converge or diverge?- Simplify term
- ∑
_{n = 1}^{∞}[(3^{n})/3] = ∑_{n = 1}^{∞}3^{n − 1}

Determine rr = 3|3| > 1, thus the series diverges

Does ∑

_{n = 1}^{∞}18( [1/3] )^{n − 1}converge or diverge?Determine rr = [1/3]|[1/3]| ≤ 1, thus the series converges

What is the sum of the first 9 terms of series a

_{n}= 7(2)^{n − 1}- Use the Sum of the first n terms equation
- S
_{n}= [(a(1 − r^{n}))/((1 − r))] - Define variables
- a = 7
- r = 2
- n = 9
- Apply equation
- S
_{n}= [(a(1 − r^{n}))/((1 − r))] - = [(7(1 − 2
^{9}))/((1 − 2))]

= 3577

What is the sum of the first 3 terms of series a

_{n}= 18( [1/3] )^{n − 1}- Use the Sum of the first n terms equation
- S
_{n}= [(a(1 − r^{n}))/((1 − r))] - Define variables
- a = 18
- r = [1/3]
- n = 3

S

_{n}= [(a(1 − r^{n}))/((1 − r))] = [(18( 1 − ( [1/3] )^{3}))/(( 1 − [1/3] ))] = 26What is the series represented by 1 + [4/5] + [16/25] + [64/125] + ...

Change terms into the contest of squares

1 + [4/5] + [16/25] + [64/125] + ... = 1 + [4/5] + [(4

1 + [4/5] + [16/25] + [64/125] + ... = 1 + [4/5] + [(4

^{2})/(5^{2})] + [(4^{3})/(5^{3})]a_{n}= ( [4/5] )^{n − 1}Does the series, 1 + [4/5] + [16/25] + [64/125] + ..., converge?

Determine rr = [4/5]|[4/5]| < 1, thus the series converges

What is the sum of the first 5 terms of the series 1 + [4/5] + [16/25] + [64/125] + ...

- Use the Sum of the first n terms equation
- S
_{n}= [(a(1 − r^{n}))/((1 − r))] - Define variables
- a = 1
- r = [4/5]
- n = 5

S

_{n}= [(a(1 − r^{n}))/((1 − r))] = [(( 1 − ( [4/5] )^{5}))/(( 1 − [4/5] ))] = 3.36What is the sum of the series 1 + [4/5] + [16/25] + [64/125] + ...

- Use the Sum of series equation
- S = [a/(1 − r)]
- Define variables
- a = 1
- r = [4/5]

S = [a/(1 − r)] = [1/(1 − [4/5])] = 5

What is the sum of the series ∑

_{n = 1}^{∞}3( [3/2] )^{1 − n}- Alter the sequence with exponent properties
- ∑
_{n = 1}^{∞}3( [3/2] )^{1 − n}= ∑_{n = 1}^{∞}3( [2/3] )^{n − 1} - Use the Sum of series equation
- S = [a/(1 − r)]
- Define variables
- a = 3
- r = [2/3]
- n = 5
- Apply equation
- S = [a/(1 − r)]
- = [3/(1 − [2/3])]

= 9

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Geometric Series

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Geometric Series 0:07
- Definition
- Expanded Form
- Convergence Rules
- Example 1: Convergence 1:22
- Example 2: Convergence 2:36
- Example 3: Convergence 3:45
- Sum of Series 5:04
- Sum of First n Terms
- Sum of Series
- Example 1: Sum of Series 5:39
- Example 2: Sum of Series 7:15
- Example 3: Sum of Series 8:24
- Example 4: Sum of Series 9:36

0 answers

Post by ANURAAG PRAKASH KAMLE on May 3, 2013

Sir i think in the example 2 the series should converge as |r|=1/2 < 1

1 answer

Last reply by: Arshin Jain

Sun May 25, 2014 2:11 PM

Post by Jinrong Shi on April 16, 2013

Question here, I believe that the sum of series forms in here can only be conducted if 0<r<1. Is that correct?