Finite Geometric Series

Geometric sequences are sequences of numbers that have a common ratio between consecutive terms. Geometric series refers to the sum of the terms of a geometric sequence.

Let's recall some basic facts about geometric sequences first. For example: $1,5,25,125,625...$ is a geometric sequence with a common ratio $r=5$

To find the nth term, $a_{n}$ of a geometric sequence, where $a_{1}$ is the first term, n is the the term number, and r is the common ratio, we can use this formula:

$a_{n}$ = $a_{1}\times r^{n-1}$

Here's another practice problem:

A geometric series is simply the summation of a geometric sequence. It can be either finite or infinite.

It seems interesting that the sum of the first $7$ terms of that last geometric series is just $1$ less than the $8$th term. This is not true for all geometric series (for example, $1 + 3 + 9 = 13 \ne 26,$ but $26 = 2 \times 13,$ which is also interesting), but let's investigate the sum of a geometric series algebraically. Using this definition for the nth term of a geometric sequence

$a_{n}$ = $a_{1}\times r^{n-1}$

We can write the sum of the first n terms of the series as

$a_{1} + a_{1}r + a_{1}r^2 + ... + a_{1}r^{n-1} = a_{1}(1 + r + r^2 + ... + r^{n-1})$

using the distributive property to factor the left-hand side. If we multiply the right-hand side by $1 - r$ we get

$a_{1}(1 + r + r^2 + ... + r^{n-1})(1 - r) \\= a_{1}((1 + r + r^2 + ... + r^{n-1}) - (r + r^2 + r^3 + ... + r^{n})) \\= a_{1}(1 - r^{n})$

(Do you see how this cancellation works?) Now all that is left to do to reclaim the original sum is to divide out the $1 - r$, as $\frac{1-r}{1-r} = 1$ if $r \ne 1$. So to find the sum $S_{n}$ of a finite geometric series with n terms, we can use the following formula:

$S_{n}$ = $\frac{a_{1}(1-r^n)}{1-r}$ where $r\ne1$

Try another problem!

The formula works with fractions and negative numbers as well:

Decimal Expansion of 1/89

Warning! Expanding 1/89 is long and painful!

The answer is $0.\overline{01123595505617977528089887640449438202247191}$

Yeah, that's pretty insane. You can in fact derive that decimal through long division, but it probably will take you at least until next week.

Instead, we are going to look at a remarkable property of this particular decimal expansion:

$1/89= \\ 0.0 \\ +0.01 \\ +0.001 \\ +0.0002 \\ +0.00003 \\ +0.000005 \\ +0.0000008 \\ +0.00000013 \\ +0.000000021 \\ +0.0000000034 \\ +0.00000000055 \\ +0.000000000089 \\ +0.0000000000144 \\ ... \\ =0.01123595505...$

Yes, the Fibonacci numbers (in decimal form) sum up to 1/89! Isn't that crazy?!

Specifically, if $F_0 = 0$, $F_1 = 1$, $F_2 = 1$, and $F_n = F_{n-1} + F_{n-2}$, then:

$\frac{1}{89} = \frac{F_0}{10} + \frac{F_1}{100} + \frac{F_2}{10^3} + \frac{F_3}{10^4} + \dots$

Let's find out why...

Let's suppose we don't yet know this is 1/89. So, just let:

$S = \frac{F_0}{10} + \frac{F_1}{100} + \frac{F_2}{10^3} + \frac{F_3}{10^4} + \dots$

Then, using the fact that each Fibonacci number is the sum of the previous two, we can rewrite this as:

$S = \frac{F_0}{10} + \frac{F_1}{100} + (\frac{F_0}{10^3} + \frac{F_1}{10^3}) + (\frac{F_1}{10^4} + \frac{F_2}{10^4}) + \dots$

We'll have lots of pairs added up together.

Now let's put everything together. We had:

$S = \frac{F_0}{10} + \frac{F_1}{100} + (\frac{F_0}{10^3} + \frac{F_1}{10^3}) + (\frac{F_1}{10^4} + \frac{F_2}{10^4}) + \dots$

Everything should have looked OK except for the rearrangement step, where we pulled out all of the first terms from each pair of parentheses, and added them together separately. You should be concerned that maybe we did something awful like this:

$$\begin{align} &(1 - 1) + (1 - 1) + (1 - 1) + \dots\\ &= (1 + 1 + 1 + \dots) - (1 + 1 + 1 + \dots)\\ &= \infty - \infty \end{align}$$

So, since all of our terms were nonnegative, it doesn't actually matter which order we added them up in.

The issue is that the sum could be infinite. Then we couldn't define it to be $S$ (it wouldn't be a real number). We need to be sure that the sum isn't infinite. It turns out that we can do this directly, by observing that the $n$-th Fibonacci number is always less than $2^n$.

To see why, observe that we can compare the first two terms like this:

$F_1 = 1 < 2^1$

$F_2 = 1 < 2^2$

Indeed, you can add up the inequalities, to find out that

$F_3 = 2 < 2^1 + 2^2$.

We now know that

$F_2 = 1 < 2^2$

$F_3 = 2 < 2^3$

This logic continues onward. You can learn more about that method by reading about Mathematical Induction, if you're curious.

Great! We've now figured out that the sum is less than the sum of a convergent Geometric Series, so it's finite, and we can indeed define its sum to be $S$.

That's the end of our journey for now. If you liked this topic, and are curious to read more on the Internet, check out these two articles:

The Remarkable Number 1/89 $\\$

The Decimal Expansion of 1/89 and Related Results

Let's suppose we want to find the sum $$S = 1+2+4+\cdots+2^{10}$$ We could just add everything up, but that would take too long! What might you do instead?

Ok, here is a pretty neat trick!

Yep, we get $$\begin{align*} 2S &= 2 + 4 + \cdots + 2^{10} + 2^{11}\\ S &= 1 + 2 + 4 + \cdots + 2^{10}\\ \text{subtracting...}\\ 2S - S &= 2^{11} - 1\\ S &= 2^{11} - 1 = 2047 \end{align*}$$

Wasn't that magical?

Yeah! This trick is quite clever. You can check out Cindy Zhang's explanation for more details and a general formula for summing geometric series.

A geometric series is a sum of numbers that follow a pattern. That pattern is that consecutive terms differ only by some common factor, denoted $r$. For instance, consider this geometric series with three terms.

$S = a + ar + ?$

Note that each term differs from the previous only by a factor of $r$ (called the common ratio), so the third term must be $r \cdot ar = ar^2$. If more terms are added on, then they similarly must be $ar^3$, $ar^4$, and so on.

In fact, there is a formula for the value of a geometric series with initial term value $a$, common ratio $r$, and $n$ terms:

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

We can use this idea in an example:

What is the value of the geometric series

$2 + \frac{2}{3} + \frac{2}{9} + \frac{2}{27} + \frac{2}{81} + \frac{2}{243}$?

There are $6$ terms, so $n = 6$ and

$S = \frac{a(1 - r^6)}{1 - r}$

Next, the starting term in the series is $2$, so $a = 2$:

$S = \frac{2(1 - r^6)}{1 - r}$

Finally, the common ratio is $\frac{1}{3}$, as $2 \cdot \frac{1}{3} = \frac{2}{3}$, $\frac{2}{3} \cdot \frac{1}{3} = \frac{2}{9}$, and so on. Thus $r = \frac{1}{3}$:

$S = \frac{2\left(1 - \left(\frac{1}{3}\right)^6\right)}{1 - \frac{1}{3}} = \frac{2\left(1 - \frac{1}{729}\right)}{\frac{2}{3}} = 2\left(\frac{728}{729}\right) \cdot \frac{3}{2} = \frac{728}{243}$

However, there are also infinite geometric series that have an infinite number of terms. In this case, the above formula has $n = \infty$. However, if $r > 1$ or $r < -1$, then $r^{\infty}$ is infinite (try some values of $r$ for yourself on your calculator; note how their magnitudes get rapidly bigger without showing any sign of letting up). That said, if $-1 < r < 1$, then the values of $r^n$ for increasing $n$ get smaller in magnitude until they reach $0$. Thus the geometric series remains finite. Under these constraints, for an infinite geometric series:

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

As before, $a$ can be any finite number.

Note there are two special cases that you must watch out for.

$r = 0$: The geometric series is not defined, as if you try to move "closer down the line to the starting term," i.e. from term $ar^n$ to $ar^{n - 1}$, then you are dividing by zero, which is not possible!

$r = -1$: This geometric series is not defined either, as you are simply oscillating the sign of the term. For instance, if $a = 1$ and $r = -1$, if the series is infinite, then you get

$S = 1 - 1 + 1 - 1 + 1 - ...$

But infinity is not a number, so it cannot be odd or even, so we do not know whether the sum equals $0$ or $-1$ when all of the numbers are [hypothetically] added up. Thus this class of geometric series is not defined either.

Exercise: Compute

$S = 6 + \frac{9}{2} + \frac{27}{8} + \frac{81}{32} + …$

Something to ponder: Why does our formula fail upon trying to compute the value of the infinite series

$0 + 0 + 0 + 0 + …$?

A geometric sequence is a recursively defined sequence of numbers where every term in the sequence (except the first) is found by multiplying the previous term by some constant $r,$ often referred to as the common ratio. For example, the sequence $$1,2,4,8,16,32,\ldots$$ is clearly geometric, as each term is the previous one multiplied by the common ratio, which, in this case, is $2$.

If we let the first term be $a$ and the common ratio be $r,$ a geometric sequence with $n$ terms can be written as: $$a, ar, ar^2, \ldots, ar^{n-2}, ar^{n-1}$$ (make sure you see why the last term is not $ar^n.$)

For more on geometric sequences, I recommend you look at the expii discussion here.

A natural question when working with geometric sequences is how calculate the sum of numbers in a geometric sequence. In mathematical terms, we are asking if there is a simple expression that calculates the sum $$a + ar + ar^2 + \ldots + ar^{n-2} + ar^{n-1}$$

Returning to our original example of $1,2,4,8,16$, we could easily compute the sum $1+2+4+8+16$ directly to get $31$. Let's think about what would happen if we had $1+2+4+8+16+\ldots+ 2^{50}$. Adding each term one by one would be incredibly inefficient. So, we would like to find and prove a simple formula. Finding the motivation for the derivation of the formula is a bit difficult, so I will provide the formula and then prove it.

The formula for a geometric series (with first term $a,$ common ratio $r,$ and $n$ terms) is: $$a + ar + ar^2 + \ldots + ar^{n-2} + ar^{n-1} = \dfrac{a(1-r^n)}{1-r}$$

Proof:

Let $$S_n = a + ar + ar^2 + \ldots + ar^{n-2} + ar^{n-1}.$$ Multiplying the sum by $r,$ we get: $$r*S_n = ar + ar^2+ar^3+\ldots+ar^{n-1}+ar^n.$$ Aha! There's that $ar^n$ term which seemed to be missing from our original sum. Now, we can subtract $r*S_n$ from $S_n,$ as that will make all the middle terms cancel out (which is how our formula looks). We have: $$\begin{align} S_n - r*S_n &= a + ar + ar^2 + \ldots + ar^{n-1} \\ &\quad \: \: -(ar + ar^2+\ldots+ar^{n-1}+ar^n) \\ &= a + (ar - ar) + (ar^2 - ar^2) + \ldots + (ar^{n-1} - ar^{n-1}) - ar^n \\ &= a - ar^n \end{align}$$ Now that we have $S_n - r*S_n = a - ar^n,$ we just need to isolate $S_n.$ Factoring both sides gives $S_n(1-r) = a(1-r^n),$ and then dividing both sides by $(1-r)$ gives $\boxed{S_n = \dfrac{a(1-r^n)}{1-r}},$ so we're done. $\blacksquare$

(Disclaimer: The upcoming sections will feature liberal usage of the term infinity which may violate technical definitions in some cases. I understand this, but I am trying to make this as simple to understand as possible for the reader.)

Now that we have a formula for a finite geometric series, we can find something similar for an infinite geometric series! Let's say our original example with powers of $2$ went on forever. We want to find the sum $$1+2+4+8+16+32+64+\ldots$$ with no end to the sequence. You're probably thinking that this sum is just $\infty,$ and you're right! Our individual terms in the series keep increasing, and therefore, so does the sum. We call a sequence where the sum approaches $\infty$ divergent. In general terms, any infinite sequence with common ratio $|\,r\,| > 1$ will approach either $\infty$ or $-\infty$. If you try plugging in some $a$ value with $|\,r\,| > 1$ and $n = \infty$ to our formula for sum of a geometric series, you will find that we will also get $\infty$ or $-\infty,$ since our numerator is $\pm\infty$ and our denominator is a constant.

Now let's try looking at a more interesting geometric sequence, one that does not sum to $\infty$. Let's look at the example $$1 + \frac12 + \frac14 + \frac18 + \frac{1}{16} + \frac{1}{32} + \ldots$$

We can try plugging this in to our formula for sum of a geometric series, as it worked with the previous infinite geometric series example. Substituting $a = 1, r = \frac12,$ and $n = \infty$ gives: $$S_n = \frac{1\left(1-\left(\frac12\right)^\infty\right)}{1-\frac12}$$ The numerator looks particularly interesting... Especially that $\left(\frac12\right)^\infty$ term. Well, as $n \rightarrow \infty,$ we know that $\left(\frac12\right)^n$ gets smaller and smaller, closer and closer to $0$. So, we can say that $\left(\frac12\right)^\infty = 0,$ making our infinite sum just $\dfrac{1(1-0)}{1-\frac12} = 2$. We have found a value not equal to $\infty$ for our infinite sum that goes on forever! In mathematical terms, our series is convergent. In general though, an infinite geometric series will only converge if $|\,r\,| < 1,$ because only then will $r^n$ become smaller and smaller as $n \rightarrow \infty$. So, the formula for an infinite geometric series with $| \, r \,| < 1$ is $\boxed{S_n = \dfrac{a}{1-r}},$ because the $1-r^n$ term just becomes $0$.

Infinite geometric series have applications in a lot of places, for example, finding the probability of winning in a game with a chance of win or loss on any given turn, or converting repeating decimals to fractions, et cetera. Infinite series (not necessarily geometric) will also be covered in great detail in calculus. This will not be your last encounter with them!

Next Lessons

In the next lesson, you will learn about the finite arithmetic series. In a few lessons, you will learn how to find the sum of an infinite series.

(Video) Geometric Series

by GVSUmath

In this video by GVSUmath, the concept of Geometric series is explained.

Finite Geometric Series

As we learned earlier, a series is the sum of all the terms in a sequence. Geometric sequences are commonly expressed as recursive sequences. Each new term is found by multiplying the previous term by a constant ratio.

If we want to find the finite geometric series, we are adding up all the terms in a geometric sequence that has a limited number of terms.

One way to do this is to add each term one by one. This works, but it can take a long time if there are many terms.

Luckily, there is a general formula that helps us find the sum much faster.

You can build on these skills to tackle word problems!