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You CAN Ace Calculus

17calculus > infinite series > geometric series

### Calculus Main Topics

Single Variable Calculus

Multi-Variable Calculus

### Tools

math tools

general learning tools

### Related Topics and Links

related topics

For a discussion of geometric sequences, see the page on sequences

A main application of a geometric series is to build power series

geometric series youtube playlist

WikiBooks - Geometric Series

Geometric Series

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

## 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.

### Search 17Calculus

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

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

solution

Practice A02

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

solution

Practice A03

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

solution

Practice A04

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

solution

Practice A05

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

solution

Practice A06

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

solution

Practice A07

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

solution

Practice A08

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

solution

Practice A09

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

solution

Practice A10

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

solution

Practice A11

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

solution

Practice A12

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

solution

Practice A13

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

solution

Practice A14

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

solution

 Level B - Intermediate

Practice B01

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

solution

Practice B02

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

solution

Practice B03

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

Express $$\displaystyle{ 0.\overline{21} }$$