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

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

Summation notation

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:

A geometric picture

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.

Sum of a 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 red subtract to zero, really simplifying the result, which we can rearrange to come up with a nice formula for the sum, S.

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

Convergence – The story of Zeno and Achilles

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 t₁, to travel half the distance to its target.

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.

More examples here

Convergence of an infinite geometric series

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

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

Applications of geometric series

1. Simple interest

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

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.

2. Is   $0.99\bar{9} = 1 \, ?$

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 $r = \frac{1}{10}:$

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

Practice problems

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