# Infinite Geometric Series

**This page originally created by Christian and Haowell. (2021)**

**An infinite geometric series is the sum of an infinite geometric sequence. These kinds of series have no last term.**

## General Knowledge of the Topic[edit]

Infinite geometric series can be written in the general expression: a_{1} + a_{1}r + a_{1}r^{2} + a_{1}r^{3} + … + a_{1}r^{∞}, where a_{1} is the first term and r is the common ratio. The common ratio is a value for which the values in a series gets consistently multiplied by. When the ratio has a magnitude greater than 1, the terms in the sequence will get larger and larger, and if you add larger and larger numbers forever, you will get infinity for an answer. Consider ^{1}⁄_{2}. A sequence might be 1,^{1}⁄_{2}, ^{1}⁄_{4},^{1}⁄_{8}, ^{1}⁄_{16}, ^{1}⁄_{32}, ^{1}⁄_{64},^{1}⁄_{128}, ^{1}⁄_{256}, ^{1}⁄_{512},^{1}⁄_{1024},^{1}⁄_{2048}, ^{1}⁄_{4096}, ^{1}⁄_{8192}, ^{1}⁄_{16384}, ^{1}⁄_{32768}, ^{1}⁄_{65536}, …. As the sequence goes on, the terms are getting smaller and smaller, slowly approaching zero.

## Finding the Sum of Infinite Geometric Series[edit]

The general formula for finding the sum of an infinite geometric series is s = ^{a1}⁄_{1-r}, where s is the sum, a_{1} is the first term of the series, and r is the common ratio. To find the common ratio, use the formula: ^{a2}⁄_{a1}, where a_{2} is the second term in the series and a_{1} is the first term in the series. Oftentimes the series may be presented in sigma notation. The general formula for this is on the right, where a = the first term, r = the common ratio, and r ≠ 0, 1.

An infinite geometric series with a definitive sum is called a convergent series, as the sequence of the sum converges closer to a particular value. If |r| > 1, the series is divergent, as the sequence diverges and the values keep getting consecutively larger, so the sum will eventually reach infinity (and thus there will be no definitive sum). For example, let's look at the series of "10 + 20 + 40 + 80 + …". In this case, the common ratio would be 2. You can see that as the series continues infinitely, the values keep getting larger and we can’t get to a definitive sum. However, in something like "10 + 5 + ^{5}⁄_{2} + ^{5}⁄_{4} + …", the common ratio is ^{1}⁄_{2}. The values in the series keep getting progressively smaller, and thus, the series will eventually add up to a definitive sum. In this case, the sum of this series is 20. How do we know that all of this is legitimate? If we look at the example above and manually calculate the terms one by one using the common ratio, we would get the following:

a_{1} = 10

a_{1} + a_{1}r = 10 + 5 = 15

a_{1} + a_{1}r + a_{1}r^{2} = 10 + 5 + ^{5}⁄_{2} = 17.5

a_{1} + a_{1}r + a_{1}r^{2} + a_{1}r^{3} = 10 + 5 + ^{5}⁄_{2} + ^{5}⁄_{4} = 18.75

Continuing this pattern, we will get the following sums:

Sum to 5 terms = 19.375

Sum to 6 terms = 19.6875

Sum to 7 terms = 19.84375

Sum to 8 terms = 19.921875

Sum to 9 terms = 19.9609375

Sum to 10 terms = 19.98046875

We could keep going and would see that the sum gets closer and closer to, but does not ever go over 20. That’s how we know the sum is 20.

Looking at the example of
We can immediately see that the first term is 2 and the common ratio is ^{4}⁄_{5}, judging by which value was substituted for which variable. Going off of this, if we use the expression, ^{a}⁄_{1-r}, we would get ^{2}⁄_{(1-4/5)} = ^{2}⁄_{1/5} = 10. Therefore, in this equation, the sum of the series is 10. If you want to find a specific term for a series (2^{nd} term, 57^{th} term, 138^{th} term, etc.) in this format, just simply substitute the variable n for the number associated with the term. For instance, using the example presented above, let’s say we want to find the 17^{th} term of the series. The expression we would use would simply be *2( ^{4}⁄_{5})*

^{n-1}=

*2(*

^{4}⁄_{5})^{17-1}=

*2(*

^{4}⁄_{5})^{16}≈ 2(0.281474977) = 0.562949953. Therefore, the 17

^{th}term is approximately 0.5629.

## Expressions in Different Notations[edit]

Sometimes you might an expression in this form:

The first thing you’ll notice is that n equals 0 instead of 1, and the common ratio is being multiplied to the power of n as opposed to n-1. That’s because in an infinite geometric series, the first term must always be multiplied to the power of 0. So if n equaled 2, the exponent would be n-2, if n equaled 3, the exponent would be n-3, and so on. The first term and common ratio do stay the same in this case, and in this example the first term is 3 and the common ratio is ^{5}⁄_{9}. However, you can’t figure out the n^{th} term just by looking at what n equals. In this case, since you need to subtract 1 from 1 to get 0, you can think of it as the n^{th} term is actually multiplied to the power of n-1. For example, if you want to find out the 56^{th} term, you need to multiply by the power of 55.

If the exponent is not equal to 0:

If you substitute n for 1, the common ratio would not be multiplied to the power of 0. This means that you can’t just use a simple formula to determine the n^{th} term. You’ll have to start with the first term and then manually multiply by the common ratio consecutively to get the term that you are looking for. Also, if the exponent is not equal to 0, the common ratio stays the same, but the first term is *not* whatever value is substituted for a. You'll have to manually calculate the first term, substituting n for whatever value is given to you underneath the sigma.

Another situation might be:

In cases like these, you must rearrange this expression so it fits with the general one. In this case, this is how you would rearrange the equation using the exponent laws:

Lets look at another expression.

We would have to rearrange this expression to fit the general one.

In cases like these, we would have to figure out the sum for each of the terms and then add them up.

Thus, the sum of this series is ^{-13}⁄_{3}. Remember that you can’t use a formula to find out the n^{th} term for this equation because the first term is not multiplied by the power of 0.

## Good Example Videos[edit]

https://www.youtube.com/watch?v=jxRqRLMliPc There are a lot of example questions here for you to practice with. The person behind this video also explains the concept clearly.

## References[edit]

*Infinite Geometric Series*, Varsity Tutors, https://www.varsitytutors.com/hotmath/hotmath_help/topics/infinite-geometric-series.html

*Infinite Geometric Series*, IntMath, https://www.intmath.com/series-binomial-theorem/3-infinite-geometric-series.php

*Infinite Geometric Series*, Khan Academy, https://www.khanacademy.org/math/ap-calculus-bc/bc-series-new/bc-series-optional/v/deriving-geometric-series-sum-formula

*Infinite Geometric Series*, Ltcconline, http://www.ltcconline.net/greenl/courses/103b/seqSeries/INFGEO.HTM