Note: MathJax breaks this page

on most mobile devices.

I am investigating a fix,

but this is low priority.

These are very rough notes on geometric series. I periodically add to or revise them as I review, relearn, and figure out what works best in my actuarial studies. Ideally, I would like these to get into a form where they would help someone who is rusty with mathematics, but it is a work in progress so the difficulty is uneven.

## Definition

A geometric series is a sum of the form

\[ a + ar + ar^2 + … + ar^k \]

in the finite case, or

\[ a + ar + ar^2 + … \]

in the infinite case. In both cases, \(a\) and \(r\) are constants.

Sidebar: Some people use the term *geometric series* to refer to only the infinite case, and refer to finite cases as *partial sums of a geometric series*. I suspect usage varies across professions, but there probably is variation within each profession as well. I don’t like to waste time on purely semantic disputes, so let’s just say it is a good idea always to be alert for context.

In summation notation the two series above are written as

\[\sum^k_{n=0} ar^n\]

and

\[\sum^{\infty}_{n=0} ar^n \]

I gradually will phase summation notation into future iterations of these notes. Summation notation helps to make things compact, and keeping things compact is useful once you have become an expert, but it is not always the best thing when you are learning, so I want to make sure there is a treatment with everything expanded first.

## Three forms of the basic finite geometric series

Let us call a series of the following form a *basic finite geometric series*:

\[1 + r + r^2 + …. + r^k \tag{Form 1}\]

where \(r\) is a real number and \(k\) is a positive integer.

If you have been away from mathematics for a while, recall the following basic rules of exponentiation:

- For all \(r\), \(r^0 = 1\).
- For all \(r\), \(r^1 = r\)

With these rules in mind, you can see that the basic finite geometric series above can be rewritten as

\[r^0 + r^1 + …. + r^k \tag{Form 2}\]

or in summation notation as

\[\sum^k_{n=0} r^n \tag{Form 3}\]

In practice, you will not see Form 2 in the wild very often: I write it out just to make clear the pattern of the exponents for those who might not be as immediately comfortable with summation notation: the exponents start with 0 and increase term by term all the way to \(k\). Moving forward, it is very important to get comfortable enough with all three of these forms to see their interchangeability without effort. In particular, you will want to be able to interchange between Forms 1 and 3 automatically.

## The sum of a basic finite geometric series

The sum of a basic finite geometric series with last term of power \(k\) is

\[\frac{1-r^{k+1}}{1-r}\]

Step by step:

Let \(B_k = 1 + r + r^2 + …. + r^k\) for some real number \(r\) and positive integer \(k\).

In form 2:

\[ B_k = r^0 + r^1 + r^2 + …. + r^k\tag{B1}\]

Multiply both sides of (B1) by *r*:

\[ \begin{align} rB_k &= r(r^0) + r(r^1) + r(r^2) + … + r(r^k) \\ &= r^{0+1} + r^{1+1} + r^{2+1} + … + r^{k+1} \\ &= r^1 + r^2 + r^3 + … + r^{k+1} \tag{B2}\end{align}\]

Compare (B1) and (B2). Notice that \(B_k\) and \(rB_k\) share all of their terms except for the first term of \(B_k\) (namely, \(r^0\)), and the last term of \(rB_k\) (namely, \(r^{k+1}\)). So if you subtract \(rB_k\) from \(B_k\), all of the middle terms will cancel:

\[ \begin{align} B_k – rB_k = r^0 &+ r^1 + r^2 + … + r^k \\ &-r^1 \,- r^2 – … \,- r^{k} – r^{k+1} \end{align} \]

This will leave you with:

\[ B_k – rB_k = r^0 – r^{k+1} \]

Factor \(B_k\) out of the left side:

\[ B_k(1-r) = r^0-r^{k+1} \]

Divide both sides by \(1-r\):

\[ B_k = \frac{r^0-r^{k+1}}{1-r} \]

Rewrite the \(r^0\) as 1 for simplicity:

\[ B_k = \frac{1-r^{k+1}}{1-r} \]

And there you have it.

## The sum of a finite geometric series in general

Let \(S_k = a + ar + ar^2 + … + ar^k \) for some real numbers \(r\) and \(a\) and positive integer *k*.

\[ S_k = ar^0 + ar^1 + ar^2 + … + ar^k \]

Factor out the \(a\):

\[S_k = a(r^0 + r^1 + r^2 + …. + r^k)\]

Notice that what we have inside the parentheses is \(B_k\):

\[S_k = aB_k\]

In the section above, we determined the formula for \(B_k\), so we can just plug that in, yielding:

\[ S_k = a\frac{1-r^{k+1}}{1-r} \]

This is the standard textbook formula for summing a finite geometric series.

This gives us a basic algorithem for determining the sum of a finite geometric series: (1) factor the first term out of the entire series, (2) apply the formula for the basic geometric series to the rest, and (3) multiply the result the first term you have factored out.

## What if there is no \(a\)?

Suppose we have a sum of the form \( r + r^2 + … + r^n \). This looks sort of like a finite geometric series, but there is no first constant \(a\)—so what do we do?

Actually, there *is* a first constant. It’s just hidden away a bit. In this case \(r\) is pulling double-duty as both constants.

You can see this by following the algorithm I described at the end of the last section: if you factor the first term (\(r\)) out of the series, we go from\[r + r^2 + …. + r^n\]

to

\[r(1 + r^1 + r^2 + …. + r^{n-1}) \]

This is the product of \(r\) with a basic geometric series with \(k = n-1\):

\[ \begin{align} rB_{n-1} &= r\frac{1-r^{(n-1)+1}}{1-r} \\&= r\frac{1-r^n}{1-r} \end{align} \]

## How about the infinite case?

The good news is that the formula for the infinite case is simpler than the formula for the finite case. The bad news is that there is a condition you have to check before using it. The other good news is that the condition is pretty simple.

Condition: An infinite series \( a + ar + ar^2 + … \) has a finite sum *if and only if* \(|r|<1\).

If the condition is satisfied, then the infinite sum is:

\[a\frac{1}{1-r} \]

That’s all there is to it. You can think of the above as what you would get if you were to plug \(k=\infty\) into the finite summation formula. For a \(|r|<1\), the limit of \(r^{k+1}\) as \(k\) approaches \(\infty\) is 0, leaving us with the above.

## What if the first term has \(r\) with a higher power than 0?

What if you run into a series like, say \( ar^2 + ar^3 + … + ar^k \)?

### Method 1

One way to approach this is just using the same basic algorithm introduced above: we divide through by the first term, giving us:

\[ ar^2 B_{k-2} = ar^2 \frac{1-r^{k-2+1}}{1-r} = ar^2 \frac{1-r^{k-1}}{1-r}\]

### Method 2

Another way to approach this is to add 0. More precisely, just add in the terms you need to complete the front end of the series and then subtract them right back out. For the example above, we add in \(a + ar\) and then subtract the same right out. That gives us:

\[ a + ar + ar^2 + … + ar^k – a – ar\]

We can now apply our formula to everything up to \(ar^k\), and we get

\[a\frac{1-r^{k+1}}{1-r} – a -ar\]

Some algebra shows that the two results are the same:

\[\begin{align} a\frac{1-r^{k+1}}{1-r} – a -ar &= a\left(\frac{1-r^{k+1}}{1-r} – 1 – r\right) \\&= a\frac{1-r^{k+1} – (1-r) – (r-r^2)}{1-r} \\&= a\frac{r^2 – r^{k+1}}{1-r} \\&=ar^2\frac{1-r^{k-1}}{1-r} \end{align}\]

### Method 3

Earlier, we noticed a possible simplifying pattern that we will now revisit. If you take the result from Method 1 and multiply the \(ar^2\) back in, you will get

\[\frac{(ar^2) – (ar^2)r^{k-1}}{1 – r} = \frac{(ar^2) – ar^{k+1}}{1-r} \]

This is the first term of your original series minus the term that would come after the last term in your original series, divided by \(1-r\). If there is any doubt about what \(r\) is, divide the second term by the first and the result will be \(r\). This pattern holds generally. For any geometric series:

\[\frac{\text{First term} – \text{Term that would come after the last term}}{1 – \frac{\text{Second term}}{\text{First term}}}\]

Applying this formula is probably the fastest and most versatile method you can use with geometric series. But don’t just memorize it: understand where it came from and how to do problems with the other methods.

## An unpleasant example

This series showed up early in a practice problem for the Society of Actuaries Financial Mathematics exam:

\[0.98^{40}v^1 + …. + 0.98^{59}v^{20}\]

The sum is most easily executed with Method 3: the first term of the series is \(0.98^{40}v^1\), the term that would come after the last one is \(0.98^{60}v^{21}\), and \(r\) is equal to \(\frac{0.98^{41}v^2}{0.98^{40}v^1} = 0.98v\), so the sum is

\[\frac{0.98^{40}v^{1} – 0.98^{60}v^{21}}{1 – \frac{0.98^{41}v^2}{0.98^{40}v^1}} = \frac{0.98^{40}v^{1} – 0.98^{60}v^{21}}{1 – 0.98v}\]

In the problem I was working on, we were indepedently given enough information to compute the value of \(v\), so getting it into the form above saved me from having to add twenty separate terms in my calculator—something one certainly does not want to have to do on an SOA exam, and probably not in general since life is short.

Now, in this problem, the presence of two different sets of increasing exponents initially confused me enough to make me uncertain about whether Method 3 still applied. This is a general problem with shortcut methods: they save time when you are sure they apply to the problem at hand; however, the greater the shortcut, the further one gets from first principles and therefore the less immediately clear it might be whether the shortcut applies to something unfamiliar. In my case, I opted for Method 1 to be more secure. If we factor the first term out of the series, we get

\[= 0.98^{40}v[1 + (0.98v)^1 + (0.98v)^2 + …. + (0.98v)^{19}] \]

Now apply the basic formula to the inside to get:

\[=0.98^{40}v\left( \frac{1 – (0.98v)^{20}}{1 – 0.98v}\right) \]

And then distribute the outside term back in to get:

\[= \frac{0.98^{40}v – 0.98^{60}v^{21}}{1 – 0.98v} \]

This is the same result as above: the extra increasing exponent does not create any weird exception to the rule. Walking through the longer method helped me to see that.