Geometric Series Summation

Geometric Series

The sum of a finite geometric series. The sum to infinity of a convergent geometric series, including the use of $|r| < 1$.

In the TMUA/ESAT we expect you to know what a geometric series [and sequence] is and recognise one when it appears. We often refer to these as Geometric Progressions [GPs] in the TMUA. You should also know and understand the standard formulae and terminology:

first term $= a$ [sometimes this is written as $ar^0$ to fit in with the formula for $u_n$]

$\text{common ratio} = r$

$\text{(nth term) } u_n = ar^{n-1}$

$\text{sum to n terms } S_n = \frac{a(1 - r^n)}{1 - r}$

$\text{sum to infinity } S_\infty = \frac{a}{1 - r} \quad \text{valid when } |r| < 1$

And you should be comfortable using $\Sigma$ notation and GPs:

$S_n = \sum_{k=1}^{n} ar^{k-1} = \sum_{k=0}^{n-1} ar^k$

Note how we have written the same expression in two different ways using the notation. (Whilst it takes some time to get used to $\Sigma$ notation, you should make some effort to be fluent using it as it will appear frequently during your higher studies and it will usually be assumed you are familiar with it and can manipulate it easily. You might eventually encounter expression with sums of sums $\Sigma \Sigma$ and even sums of sums of sums $\Sigma \Sigma \Sigma$ etc.)

There are some additional useful formulae/techniques you should keep in mind:

The ratio of adjacent terms is fixed: $\frac{u_{n+1}}{u_n} = r$ so $\frac{u_{n+1}}{u_n} = \frac{u_n}{u_{n-1}}$ etc.

You can derive new progressions from a given GP. Given

$S_n = a + ar + ar^2 + ar^3 + ar^4 + ar^5 \dots \dots \dots$

then we can replace $r$ by $-r$ to get

$S_{n}(-r) = a - ar + ar^{2} - ar^{3} + ar^{4} - ar^{5} \dots \dots \dots$

and we note that $S_{n}(-r)$ is also a GP with first term $a$ and common ratio $-r$ and

$S_{\infty} = \frac{a}{1 + r}$

Using this, we can also create other GPs. Consider this progression:

$\frac{1}{2} \left( S_{n} + S_{n}(-r) \right) = a + ar^{2} + ar^{4} + ar^{6} \dots \dots \dots$

Which is a GP with $u_{1} = a$ and a common ratio of $r^{2}$. What is its sum to infinity?

You can also explore $\frac{1}{2} \left( S_{n} - S_{n}(-r) \right)$.

And we can generate new GPs by raising each term in $S_{n}$ to a power. Here are some examples:

Example 1

Square every term:

$S_{n}(squared) = a^{2} + a^{2}r^{2} + a^{2}r^{4} + a^{2}r^{6} + \dots \dots \dots$

So $S_{n}(squared)$ is a GP with $u_{1} = a^{2}$ and a common ratio of $r^{2}$.

What is $S_{\infty}$ (squared)?

Example 2

Raise every term to power of $k$ [we will assume $k$ is a positive integer]:

$S_n(\text{power is } k) = a^k + a^k r^k + a^k r^{2k} + a^k r^{3k} + \cdots \dots \dots$

Verify that $S_n(\text{power is } k)$ is also a GP. What can you say about $S_\infty(\text{power is } k)$ when $|r| < 1$? What happens if $k$ is any positive number? And what about if $k < 0$ and $0 < r < 1$?

We can also look at sum of part of a GP. Given

$S_k = a + ar + ar^2 + ar^3 + ar^4 + \cdots + ar^{k-1}$

find an expression for $ar^m + ar^{m+1} + \cdots + ar^n$ [where we assume $n > m$].

When exploring these sorts of expression, you have to be very careful with the ends of the sequence as it can be easy to write the wrong thing when simplifying. For instance, it is tempting to write this expression as $S_n - S_m$ but that would be wrong [why?]. And also, for instance, we might ask how many terms are there in the sum $ar^m + ar^{m+1} + \cdots + ar^n$? Is it $n - m$ terms? (No! It is $n - m + 1$ terms. Can you see why? This is an example of the fence post error – it is very easy to make this error, so explore what it is and think about how you would avoid making it when answering questions.)

There are lots of ways of tackling this sum and how you approach it might depend on how the problem you are dealing with is set up:

Method 1

Treat is as the difference of two sums:

$ar^m + ar^{m+1} + \cdots + ar^n = S_{n+1} - S_m$

Note the $n + 1$ in $S_{n+1}$.

Method 2

Factorise out $r^m$ and treat what is in the bracket as a simple GP with $u_1 = a$ and common ratio $r$ and number of terms $n - m + 1$:

$ar^{m} + ar^{m+1} + \cdots + ar^{n} = r^{m}(a + ar + \cdots + ar^{n-m})$

Or even better, factorise out $ar^{m}$:

$ar^{m} + ar^{m+1} + \cdots + ar^{n} = ar^{m}(1 + r + \cdots + r^{n-m})$

Method 3

Treat it as a new GP with first term $ar^{m}$ and common ratio $r$ and number of terms $n - m + 1$:

$ar^{m} + ar^{m+1} + \cdots + ar^{n} = ar^{m} + (ar^{m})r + (ar^{m})r^{2} + \ldots + (ar^{m})r^{n-m}$

Exercise

For each of the above methods, work out the sum and verify all three methods give you the same answer. This is more an exercise in being careful than anything else – you should not skip this exercise thinking you can do it easily!

There are lots of questions on geometric sequences in the TMUA/ESAT past papers. We recommend you work through the past papers first [initially under timed conditions] and if you get stuck, study the detailed worked answers we have supplied for each TMUA/ESAT paper.