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10.2 Working with Geometric Series

1 min readjune 7, 2020

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Welcome to the first of many convergence tests that you will learn called the geometric series test! But before we start, we’ll go over what exactly a geometric series is and the significance of having one!

If you would like a review of what a convergent vs. divergent infinite series is, check out our 10.1 guide: Defining Convergent and Divergent Infinite Series!


🔺 What is a Geometric Series?

A geometric series is a series that follows the standard format as shown below. Pay attention to the difference in where the sequence starts! The first series starts from 0 and the second series starts from 1!

sn=n=0arns_n = \sum_{n=0}^{\infty} a \cdot r^n
sn=n=1arn1s_n = \sum_{n=1}^{\infty} a \cdot r^{n-1}

The definition of a geometric series is a series with a constant ratio between successive terms. Therefore aa is your initial term and rr is the ratio between any two consecutive terms!


👷 Building a Geometric Series from a Given Sequence

With this general form of a geometric series, you can construct your own with a given sequence!

✏️ Writing a Geometric Series: Example 1

Find the geometric series for the following sequence:

27,9,3,1,13,19,...27,9,3,1,\frac{1}{3},\frac{1}{9},...

🔎 Step 1: Determine the values of aa and rr

Since aa is the initial value, a=27a=27. Now to determine rr, take any two successive terms and determine what factor the first is multiplied by to get the second term and that is your value for rr! Since 913=39 \cdot \frac{1}{3} = 3, r=13r=\frac{1}{3}.

🟰 Step 2: Plug in your values to your geometric series

sn=n=027(13)ns_n = \sum_{n=0}^{\infty} 27 \cdot (\frac{1}{3})^n
sn=n=127(13)n1s_n = \sum_{n=1}^{\infty} 27 \cdot (\frac{1}{3})^{n-1}

Perfect! Let’s try one more before we get into the test itself!

✏️ Writing a Geometric Series: Example 2

Find the geometric series for the following sequence:

2,6,18,54,...2,-6,18,-54,...

Now it’s your turn! 😀 Follow the same steps from above to set up your infinite geometric series! Check your answers below to see if you got it right! ✅

🔎 Step 1: Determine the values of aa and rr

a=2r=3a=2 \\ r=-3

🟰 Step 2: Plug in your values to your geometric series

sn=n=02(3)ns_n = \sum_{n=0}^{\infty} 2 \cdot (-3)^n
sn=n=12(3)n1s_n = \sum_{n=1}^{\infty} 2\cdot (-3)^{n-1}

Later on, we’ll come back to these examples to determine if they are converging or diverging!


🤓 The Geometric Series Test

The geometric series test is a theorem that states:

A geometric series converges if 0<r<10<|r|<1 and diverges if r1|r|\geq 1 , where aa is the first term of the series and rr is the common ratio between consecutive terms.

To determine where the series converges (also called the sum of the series), the following equation is used:

n=1arn1=n=0arn=a1r\sum_{n=1}^{\infty} a\cdot r^{n-1} = \sum_{n=0}^{\infty} a \cdot r^n = \frac{a}{1-r}

Now let’s go back to our first examples and determine whether they are convergent or divergent!

💭Geometric Test: Example 1 (continued)

Here’s what we got when determined the series above:

sn=n=027(13)ns_n = \sum_{n=0}^{\infty} 27 \cdot (\frac{1}{3})^n
sn=n=127(13)n1s_n = \sum_{n=1}^{\infty} 27 \cdot (\frac{1}{3})^{n-1}

We determined that r=13r=\frac{1}{3} for the first example. Therefore, according to the geometric series test, this series is convergent!

Normally, the question will ask you to find the sum of the series, so let’s use the given equation and determine the partial sum.

27113=40.5\frac{27}{1-\frac{1}{3}} = 40.5

Your final reasoning for a question like this would be “This sum of the series is 40.5 because of the geometric series test.” All you need to include in your math is the series written geometrically and solving for where the series converges! 😀

💭Geometric Test: Example 2 (continued)

What about this one?!

sn=n=02(3)ns_n = \sum_{n=0}^{\infty} 2 \cdot (-3)^n
sn=n=12(3)n1s_n = \sum_{n=1}^{\infty} 2\cdot (-3)^{n-1}

Don’t let the negative sign in rr fool you! Follow the same steps as seen in Example 1. Since r=3r=-3, r1|r| \geq 1 and therefore the series diverges! All you have to mention is that it diverges because of the geometric series test!


📕 Closing

So hey, you’ve finished your first convergence test! Keep practicing and you’ll eventually be a total pro at infinite geometric series! Onward and upward! 🌟

10.2 Working with Geometric Series

1 min readjune 7, 2020

Athena_Codes

Athena_Codes

Athena_Codes

Athena_Codes

Attend a live cram event

Review all units live with expert teachers & students

Welcome to the first of many convergence tests that you will learn called the geometric series test! But before we start, we’ll go over what exactly a geometric series is and the significance of having one!

If you would like a review of what a convergent vs. divergent infinite series is, check out our 10.1 guide: Defining Convergent and Divergent Infinite Series!


🔺 What is a Geometric Series?

A geometric series is a series that follows the standard format as shown below. Pay attention to the difference in where the sequence starts! The first series starts from 0 and the second series starts from 1!

sn=n=0arns_n = \sum_{n=0}^{\infty} a \cdot r^n
sn=n=1arn1s_n = \sum_{n=1}^{\infty} a \cdot r^{n-1}

The definition of a geometric series is a series with a constant ratio between successive terms. Therefore aa is your initial term and rr is the ratio between any two consecutive terms!


👷 Building a Geometric Series from a Given Sequence

With this general form of a geometric series, you can construct your own with a given sequence!

✏️ Writing a Geometric Series: Example 1

Find the geometric series for the following sequence:

27,9,3,1,13,19,...27,9,3,1,\frac{1}{3},\frac{1}{9},...

🔎 Step 1: Determine the values of aa and rr

Since aa is the initial value, a=27a=27. Now to determine rr, take any two successive terms and determine what factor the first is multiplied by to get the second term and that is your value for rr! Since 913=39 \cdot \frac{1}{3} = 3, r=13r=\frac{1}{3}.

🟰 Step 2: Plug in your values to your geometric series

sn=n=027(13)ns_n = \sum_{n=0}^{\infty} 27 \cdot (\frac{1}{3})^n
sn=n=127(13)n1s_n = \sum_{n=1}^{\infty} 27 \cdot (\frac{1}{3})^{n-1}

Perfect! Let’s try one more before we get into the test itself!

✏️ Writing a Geometric Series: Example 2

Find the geometric series for the following sequence:

2,6,18,54,...2,-6,18,-54,...

Now it’s your turn! 😀 Follow the same steps from above to set up your infinite geometric series! Check your answers below to see if you got it right! ✅

🔎 Step 1: Determine the values of aa and rr

a=2r=3a=2 \\ r=-3

🟰 Step 2: Plug in your values to your geometric series

sn=n=02(3)ns_n = \sum_{n=0}^{\infty} 2 \cdot (-3)^n
sn=n=12(3)n1s_n = \sum_{n=1}^{\infty} 2\cdot (-3)^{n-1}

Later on, we’ll come back to these examples to determine if they are converging or diverging!


🤓 The Geometric Series Test

The geometric series test is a theorem that states:

A geometric series converges if 0<r<10<|r|<1 and diverges if r1|r|\geq 1 , where aa is the first term of the series and rr is the common ratio between consecutive terms.

To determine where the series converges (also called the sum of the series), the following equation is used:

n=1arn1=n=0arn=a1r\sum_{n=1}^{\infty} a\cdot r^{n-1} = \sum_{n=0}^{\infty} a \cdot r^n = \frac{a}{1-r}

Now let’s go back to our first examples and determine whether they are convergent or divergent!

💭Geometric Test: Example 1 (continued)

Here’s what we got when determined the series above:

sn=n=027(13)ns_n = \sum_{n=0}^{\infty} 27 \cdot (\frac{1}{3})^n
sn=n=127(13)n1s_n = \sum_{n=1}^{\infty} 27 \cdot (\frac{1}{3})^{n-1}

We determined that r=13r=\frac{1}{3} for the first example. Therefore, according to the geometric series test, this series is convergent!

Normally, the question will ask you to find the sum of the series, so let’s use the given equation and determine the partial sum.

27113=40.5\frac{27}{1-\frac{1}{3}} = 40.5

Your final reasoning for a question like this would be “This sum of the series is 40.5 because of the geometric series test.” All you need to include in your math is the series written geometrically and solving for where the series converges! 😀

💭Geometric Test: Example 2 (continued)

What about this one?!

sn=n=02(3)ns_n = \sum_{n=0}^{\infty} 2 \cdot (-3)^n
sn=n=12(3)n1s_n = \sum_{n=1}^{\infty} 2\cdot (-3)^{n-1}

Don’t let the negative sign in rr fool you! Follow the same steps as seen in Example 1. Since r=3r=-3, r1|r| \geq 1 and therefore the series diverges! All you have to mention is that it diverges because of the geometric series test!


📕 Closing

So hey, you’ve finished your first convergence test! Keep practicing and you’ll eventually be a total pro at infinite geometric series! Onward and upward! 🌟



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© 2024 Fiveable Inc. All rights reserved.

AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.