This page covers geometric series. The infinite series page is a general introduction to series, and geometric series are related to the other kinds of infinite series linked here.

Make sure you are familiar with all of the material on the infinite series introduction page before you work through geometric series.

A geometric series is a sum of terms (the number of terms can be finite or infinite) in which each pair of adjacent terms has a constant ratio. For the series,

$$S = a_1 + a_2 + a_3 + a_4 + \dots,$$

the ratios

$$\frac{a_2}{a_1}, \: \frac{a_3}{a_2}, \: \frac{a_4}{a_3} \dots$$

are all equal, and we call the constant ratio $r.$

We can write series like this in a couple of ways, one is

$$S = a_o + a_o r + a_o r^2 + \dots + a_o r^n = \sum_{k=0}^{n} a_o r^k$$

$$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots = \sum_{n=0}^{\infty} \, \left( \frac{1}{2} \right)^n$$

is a geometric series with $r = \frac{1}{2}$ and $a_o = 1.$

$$1 + 0.1 + 0.01 + 0.001 + ... = \sum_{n=0}^{\infty} \, \left( \frac{1}{10} \right)^n$$

is a geometric series with $r = \frac{1}{10}$ and $a_o = 1.$

$$\frac{3}{3} + \frac{3}{4} + \frac{3}{8} + \frac{3}{16} + \frac{3}{32} = \sum_{k=0}^4 \; 3 \left( \frac{1}{2} \right)^k$$

is a geometric series with five terms, and $r = \frac{1}{2}$ and $a_o = 3.$

Successive terms of a geometric series have a constant ratio.

A geometric series can be written as

$$S = a_o + a_o r + a_o r^2 + \dots + a_o r^n = \sum_{k=0}^{n} a_o r^k,$$

where $ r $ is the constant ratio and $a_o$ is a constant.

In the examples above we used summation notation. The large Greek letter sigma, Σ, generally means "sum" or "summation" in math. Here's an example that we can take apart and analyze:

$$\sum_{n=0}^{4} \left( \frac{1}{2} \right)^n$$

In Σ notation, the expression below gives the counting or "index" variable, often one of the letters, i, j, k, m, or n, and its starting value. The number above the Σ is the final value of the index, and we assume that the index counts by 1 unit. So this sum has five terms, labeled 0-4, and in each, *n* has its value for that iteration of the sum. This summation means:

$$= \left( \frac{1}{2} \right)^0 + \left( \frac{1}{2} \right)^1 + \left( \frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^3 + \left( \frac{1}{2} \right)^4$$

$$= 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}.$$

With a little practice, you'll get used to this notation, both in translating summation notation to the terms of a series, and in summarizing a bunch of terms with some nice compact summation notation.

The sum

$$\sum_{n=0}^{\infty} \, \left( \frac{1}{2} \right)^n$$

simply means that n runs, in steps of 1 unit, from 0 to infinity. This is called an **infinite series**.

Here is another representation of the numbers in the series we discussed in Zeno's paradox, a geometric series:

Each successive black section of the square above is a successive power of the **constant ratio** ^{1}/_{2}. You can see how rapidly the terms of the series become very small; this is a convergent **sequence** of terms and would form a convergent series — it would have an upper bound.

Here is the algebraic representation of this geometric series.

To derive a formula for the sum of a geometric series, consider a series, **S**, with a constant ratio of **r**, then write **r·S** by multiplying **r** by every term. (Notice that here we've begun our sum with an index of 1 and adjusted the exponent to k = 1, but the resulting series as the same as those above. The notation is a little flexible.)

$$ \begin{align} S = \sum_{k=1}^n \, a r^{k-1} &= a + ar + ar^2 + ar^3 + \dots + ar^{n-2} + ar^{n-1} \\[5pt] rS &= \phantom{000} \, ar + ar^2 + ar^3 + \dots + ar^{n-2} + ar^{n-1} + ar^n \end{align}$$

Now subtract the two lines. Notice that all of the terms in

$$ \begin{align} S = \sum_{k=1}^n \, a r^{k-1} &= a + \color{red}{ar + ar^2 + ar^3 + \dots + ar^{n-2} + ar^{n-1}} \\[2pt] rS &= \phantom{000} \, \color{red}{ar + ar^2 + ar^3 + \dots + ar^{n-2} + ar^{n-1}} + ar^n \\[4pt] \hline S = rS &= a - a r^n \\[5pt] S(1 - r) &= a(1 - r^n) \\[8pt] S &= \frac{a(1 - r^n)}{1 - r} \end{align}$$

The sum of the first m terms of a geometric series is

$$S = \sum_{n = 1}^m \, a \, r^{n-1} = \frac{a (1 - r^m)}{1 - r}$$

Some infinite series, including geometric series, **converge** to a limiting value, while others do not. That's an odd notion. How can we continue to add finite values (albeit getting smaller and smaller) to a series and have it not grow forever? Here's one way to think about it.

Consider the story of Zeno and Achilles: The two have an argument and Achilles tells Zeno he's going to have to kill him – shoot him with an arrow. But Zeno says "Go ahead, you won't kill me!"

**Achilles**: "But that's ridiculous, I've shot and killed many men with my bow, and I'll kill you, too!"

**Zeno**: "No, my fine, hulking friend, your arrow will never reach me!"

**Achilles**: "Explain yourself, Sir!"

**Zeno explains** that when Achilles releases his arrow, they would have to agree that it will take a certain amount of time, let's call it t_{1}, to travel half the distance to its target.

Afterward it will take another, smaller but still finite, time, t_{2}, to travel half of the remaining distance.

It will take a time t_{3} (smaller still, but measurable!) to travel half of the remaining distance, and so on, **infinitely**. Now, Zeno argues, if we add an infinite number of finite times, even though some are quite small, we end up with an infinite time - the arrow will never reach me!

Achilles fires his arrow and kills poor Zeno dead as a doorknob. The series, each successive term (a time) half of the previous, **converges** to exactly the time it would take for an arrow to travel between Achilles and Zeno.

*A brilliant student of mine once noted that Achilles could just aim well behind Zeno and put the arrow through him on its way to the first halfway point.*

Some series converge to a finite limit, and some don't (they **diverge**). On the right are the first several terms of the series

$$\sum_{n=0}^{\infty} \, \frac{1}{n!},$$

This series actually converges to the number *e*, the base of all continuously-growing exponential functions. Notice that successive terms only modify places further and further to the right of the decimal. That's what convergence means: The series approaches a finite limit as the number of terms grows.

Some series don't converge at all. Thos sums just keep growing without bound. We call this **divergence**, the opposite of convergence.

Convergent series are the most important kind. Series are usually used to find an alternate route to the value of a function that's difficult to find directly — we want convergence to some solution. We generally aren't too concerned with series that grow without bound.

Looking at the examples of geometric series shown so far, it's not too difficult to see that if the constant ratio is less than one, then the successive terms of an *infinite* series will get smaller and the series will converge to a limit. If the constant ratio is one or more, the terms will either stay the same size or get larger, so the sum of the series will grow without bound.

Let's look at that again: go back to the formula we derived above for the sum of a geometric series with a finite number of terms:

$$s = \frac{a(1 - r^n)}{1 - r}$$

Now if we take the limit of this function as $n \rightarrow \infty,$ then we have

$$s = \lim_{n \rightarrow \infty} \, \frac{a(1 - r^n)}{1 - r}$$

Now if r ≥ 1, then this series will diverge, either because (in the case that r = 1) it blows up, or because if r > 1, then r^{n} → ∞ as $n \rightarrow \infty.$

Finally, if $\lim_{n \rightarrow \infty} \, r^n = 0$ when $r \rightarrow \infty,$ then the sum of an infinite, convergent geometric series is just

$$S = \frac{a}{1 - r},$$

pretty simple!

If r < 1, then the series will converge to a finite limit (upper or lower bound), and the sum of the series is

$$S = \frac{a}{1 - r},$$

If r ≥ 1, then the series will grow without bound (diverge).

Let's say we have a bank account that pays an interest rate of **r** per year (annual interest). The amounts per year are given in the table below. If our initial investment is **a _{o}** in year zero, then the interest earned (first row of the table) by the end of that year is $a_o r.$ The amount in the account at the beginning of year 1 is $a_o$ plus the interest from the previous year, $a_o r,$ where $a_o + a_o r = a_o (1 + r).$ At the end of that year, we take that amount and multiply it by $r$ to get the new balance to begin year

Time | Amount | Interest |

$t = 0$ | $a_o$ | $a_o r$ |

$t = 1$ | $a_o (1 + r)$ | $a_o r (1 + r)$ |

$t = 2$ | $a_o (1 + r)^2$ | $a_o r (1 + r)^2$ |

$t = 3$ | $a_o (1 + r)^3$ | $a_o r (1 + r)^3$ |

$t = 4$ | $a_o (1 + r)^4$ | $a_o r (1 + r)^4$ |

Now there are two ways we can calculate the amount present in the account at some time, t. First, we can use the simple growth formula,

$$A(t) = a_o (1 + r)^t,$$

and calculate the amount directly.

The second is to notice that the third column, containing the amount *added* each year, is a geometric series with a constant ratio of $(1 + r).$

$$A(T) = a_o + \sum_{t=0}^T \, a_o r (1 + r)^t.$$

Using our formula for the sum of the first few terms of a geometric series, we have

$$ \begin{align} A(T) &= a_o + \frac{a_o r (1 - (1 + r)^t)}{1 - (1 + r)} \\[5pt] &= a_o + \frac{a_o r (1 - (1 + r)^t)}{-r} \\[5pt] &= a_o + -a_o (1 - (1 + r)^t) \end{align}$$

As an example to show that these are identical representations, let's let $r = 0.1,$ $a_o = 100,$ and $t = 5.$

The simple growth formula gives

$$ \begin{align} A(5) &= 100 (1 + 0.1)^5 \\[5pt] &= 161.05 \end{align}$$

The series representation gives

$$ \begin{align} A(5) &= 100 - 100(1 - (1 + 0.1)^5) \\[5pt] &= 161.05 \end{align}$$

This is kind of a fun question. Does the number 0.999999 ... represent (or converge to a value of 1 ? Well, we can write the number as

$$0.9999 = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \dots$$

Now this is a geometric series with $r = \frac{1}{10}:$

$$0.999 = 9 \, \sum_{n = 1}^{\infty} \, \left( \frac{1}{10} \right)^n$$

In order to find the sum, let's rewrite it slightly as

$$S = 9 \left( \sum_{n = 0}^{\infty} \, \left( \frac{1}{10} \right)^n - 1 \right),$$

where we've simply added and subtracted 1 to get a zero start value for our index variable n, so we can use our geometric series sum formula (infinite series):

The sum is

$$S = 9 \left( \frac{1}{1 - \frac{1}{10}} - 1 \right) = 1$$

So 0.9999 ... does indeed converge to 1.

For each of the following, determine whether the series converges or diverges. If it converges, find its sum:

1. |
$$1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots$$ ## SolutionThis is a convergent geometric series with $r = \frac{1}{3} \lt 1$ and $a_o = 1.$ Its sum is: $$\sum_{n=0}^{\infty} \, \left( \frac{1}{3} \right) = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2}$$ |

2. |
$$2 + 1 + \frac{1}{2} + \frac{1}{4} + \dots$$ ## SolutionFirst, we can recognize that this is a gemetric series plus a constant: $$S = 2 + \sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n$$ The sum is a convergent geometric series with $r = \frac{1}{2} \lt 1$ and $a_o = 1.$ Its sum is: $$ \begin{align} S &= 2 + \sum_{n=0}^{\infty} \, \left( \frac{1}{2} \right)^n \\[5pt] &= 2 + \frac{1}{1 - \frac{1}{2}} = 2 + 2 = 4 \end{align}$$ |

3. |
$$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots$$ ## SolutionThis series can be rewritten like this: $$S = -1 + \sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n$$ The sum part is a convergent geometric series with $r = \frac{1}{2} \lt 1$ and $a_o = 1.$ Its sum is: $$ \begin{align} S &= -1 + \frac{1}{1 - \frac{1}{2}} \\[5pt] &= -1 + 2 = 1 \end{align}$$ |

4. |
$$1 + 1.01 + (1.01)^2 + (1.01)^3 + \dots$$ ## SolutionThis is a |

5. |
$$\sum_{n = 1}^{\infty} \left( \frac{\pi}{4} \right)^n$$ ## SolutionThis is convergent geometric series with $r = \frac{\pi}{4} \lt 1$ and $a_o = 1.$ The sum is $$ \begin{align} S &= \frac{1}{1 - \frac{\pi}{4}}\\[5pt] &= \frac{1}{\frac{4 - \pi}{4}} \\[5pt] &= \frac{4}{4 - \pi} \approx 4.66 \end{align}$$ |

6. |
$$\sum_{n = 1}^{\infty} \; 2^{n - 1}$$ ## SolutionWe can rewrite this sum as $$\sum_{n=1}^{\infty} \, 2^{n-1} = \sum_{n=1}^{\infty} \, 2^n 2^{-1} = \frac{1}{2} \sum_{n=1}^{\infty} \, 2^n$$ This is a divergent geometric series with $r = 2 \gt 1,$ so we can't find the sum of an infinite number of terms. |

7. |
$$\frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \frac{4}{5} + \dots$$ ## SolutionIf we examine the ratios between the first two pairs of terms we get $$\frac{2/3}{1/2} = 3, \: \frac{3/4}{2/3} = 2$$ This is |

8. |
$$\sum_{n = 1}^{\infty} \; 8^{-n}$$ ## SolutionFor clarity, we can rewrite this series as $$S = \sum_{n=1}^{\infty} \left( \frac{1}{8} \right)^n = -1 + \sum_{n=0}^{\infty} \left( \frac{1}{8} \right)^n$$ Its sum is $$ \begin{align} S &= -1 + \frac{1}{1 - \frac{1}{8}} \\[5pt] &= -1 + \frac{1}{7/8} \\[5pt] &= -1 + \frac{8}{7} = \frac{1}{7} \end{align}$$ |

9. |
$$\sum_{n = 0}^{\infty} \left( \frac{1}{1 + x^2} \right)^n$$ ## SolutionThis series will converge as long as $x \gt 0.$ Its sum is $$ \begin{align} S &= \frac{1}{1 - \frac{1}{1 + x^2}} \\[5pt] &= \frac{1}{\frac{1 + x^2 - 1}{1 + x^2}} \\[5pt] &= \frac{1 + x^2}{x} = \frac{1}{x^2} + 1 \end{align}$$ |

10. |
$$1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt[3]{3}} + \frac{1}{\sqrt[4]{4}} + \dots$$ ## SolutionFor any series like this, we should probably check some ratios to see if it's actually a geometric series. The first few ratios are $$ \begin{align} &\frac{1}{2^{1/2}}, \; \frac{2^{1/2}}{3^{1/3}}, \; \frac{3^{1/3}}{4^{1/4}} \\[5pt] &\approx 0.707, \; 0.980, \; 1.02 \end{align}$$ The ratio is not constant, so this is not a geometric series. |

11. | A ball that rebounds 65% of the distance fallen is dropped from a height of 12 ft. How far does the ball travel? | |

12. | Suppose that 75 cents of every dollar spent in the U.S. is spent again in the U.S. If the federal government pumps an extra 1 billion dollars into the economy, how much spending occurs as a result? |

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