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17calculus > infinite series > geometric series

Geometric Series

on this page: ► geometric series convergence     ► how to use geometric series     ► finite geometric series     ► additional topics

Geometric Series are an important type of series that you will come across while studying infinite series. This series type is unusual because not only can you easily tell whether a geometric series converges or diverges but, if it converges, you can calculate exactly what it converges to. This is extremely unusual for an infinite series. Let's break down what this theorem is saying and how we can use it.

Geometric Series Convergence

A series in the form \( \displaystyle{ \sum_{n=0}^{\infty}{r^n}} \) is called a Geometric Series with ratio \(r\).

\( 0 \lt \abs{r} \lt 1 \)

\( \abs{r} \geq 1 \)

converges

diverges

If the series converges, it converges to \(\displaystyle{\frac{1}{1-r}} \).

This is usually written \(\displaystyle{ \sum_{n=0}^{\infty}{r^n} = \frac{1}{1-r}, ~~ 0 \lt \abs{r} \lt 1 }\).

Quick Summary

used to prove convergence

yes

used to prove divergence

yes

can be inconclusive

no

can find convergence value

yes

useful for finding power series

How To Use The Geometric Series

A geometric series looks like this \(\displaystyle{ \sum_{n=0}^{\infty}{ r^n } }\) where r is an expression of some sort, not containing n.

If you can get your series into this form using algebra, then \(r\) will tell you whether the series converges or diverges.
If \( \abs{r} \geq 1 \), then the series diverges.
If \( \abs{r} < 1, \) then the series converges and it converges to \( \displaystyle{\frac{1}{1-r}} \)

Notes

1. r is called the ratio.
2. The absolute values on r to determine convergence or divergence are absolutely critical. Do not drop them unless you are sure that r is positive all the time. If you are not sure, keeping them is always correct.
3. Watch out! - - Be careful to notice that the series given above starts with index \(n=0\). When determining the convergence value, make sure you take that into account and adjust your series to exactly match this one, including the starting index value of zero.

Finite Geometric Series

In the following discussion, we are assuming that \( 0 < r < 1 \).
The equation for the value of a finite geometric series is
\(\displaystyle{ \sum_{n=0}^{k}{r^n} = \frac{1-r^{k+1}}{1-r} ~~~~~ (1) }\)
where \(k\) is a finite positive integer.

Practice 1

Show that the above formula holds for \(k=1, k=2\) and \(k=3\).

solution

Derivation

In this section, we will derive equation (1).
Let \(\displaystyle{ S_n = \sum_{k=0}^{n}{r^k} = }\) \(1+r+r^2+r^3+ . . . + r^n\)
Multiplying this by \(r\) gives us \(rS_n = r + r^2 + r^3 + . . . +r^n + r^{n+1} \)
Now we subtract \(rS_n\) from \(S_n\) to get \(S_n - rS_n = (1+ r + r^2 + r^3 + . . . +r^n) \) \(-\) \( (r+r^2+r^3+ . . . + r^n + r^{n+1} )\)
Looking closely at the terms on the right side of the equal sign, we can see that all the terms cancel except for \(1-r^{n+1}\). So now we have \(S_n-rS_n = 1-r^{n+1}\). Now we solve for \(S_n\).
\(\begin{array}{rcl} S_n-rS_n & = & 1-r^{n+1} \\ S_n(1-r) & = & 1-r^{n+1} \\ S_n & = & \displaystyle{ \frac{1-r^{n+1}}{1-r} } \end{array}\)

\(\displaystyle{ \sum_{k=0}^{n}{r^k} = \frac{1-r^{k+1}}{1-r} }\) [qed]

Additional Geometric Series Topics

video: Financial Application of Geometric Series

Financial Application of Geometric Series

This video shows a geometric series applied to a financial equation. The application of calculus to finances is not covered on this site. But you may find this video helpful if you are in business calculus.



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Practice Problems

Instructions - - Unless otherwise instructed, for each of the following series
1. determine whether it converges or diverges, using the geometric series test, if possible.
2. If it converges, determine what it converges to (if possible). Give your answers in exact form.

Level A - Basic

Practice A01

\(\displaystyle{\sum_{n=0}^{\infty}{\frac{3}{4^n}}}\)

answer

solution

Practice A02

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^n} } }\)

answer

solution

Practice A03

\(\displaystyle{\sum_{n=1}^{\infty}{\left[\frac{1+\sqrt{5}}{2}\right]^n}}\)

answer

solution

Practice A04

Express \(\displaystyle{ 1+0.4+0.16+0.064+ . . . }\) using sigma notation and determine convergence or divergence.

answer

solution

Practice A05

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\pi^n }{ 3^{n+2}} } }\)

answer

solution

Practice A06

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{ 3^n+2^n }{ 6^n } } }\)

answer

solution

Practice A07

Express this sum using sigma notation
\(\displaystyle{ 1+0.1+0.01+0.001+0.0001+ . . . }\)

answer

solution

Practice A08

\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{ 1 }{ 5^n } } }\)

answer

solution

Practice A09

Express this sum using sigma notation
\(\displaystyle{ 2/13 - 4/13^2 + 8/13^3 - 16/13^4 + . . . }\)

answer

solution

Practice A10

\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{3^n}{4^n} + \frac{2}{7^n} \right] } }\)

answer

solution

Practice A11

\(\displaystyle{ \sum_{k=0}^{\infty}{ \left( \frac{2}{3} \right)^k } }\)

answer

solution

Practice A12

\(\displaystyle{ 1 + e^{-1} + }\)
\(\displaystyle{ e^{-2} + e^{-3} + . . . }\)

answer

solution

Practice A13

\(\displaystyle{ \frac{3}{2} - \frac{3}{4} + \frac{3}{8} - \frac{3}{16} + . . . }\)

answer

solution

Practice A14

\(\displaystyle{ 1 - \frac{1}{5} + \left( \frac{1}{5} \right)^2 - \left( \frac{1}{5} \right)^3 + . . . }\)

answer

solution


Level B - Intermediate

Practice B01

\(\displaystyle{\sum_{n=1}^{\infty}{\frac{(-1)^n2^{n-1}}{3^n}}}\)

answer

solution

Practice B02

\(\displaystyle{ \sum_{n=3}^{\infty}{ 5 (2/3)^{n-1} } }\)

answer

solution

Practice B03

\(\displaystyle{ \sum_{n=3}^{\infty}{ \frac{ \pi^{n+1} }{ 6^n } } }\)

answer

solution

Practice B04

Express \(\displaystyle{ 0.\overline{21} }\)
as a fraction.

answer

solution

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