We will look at geometric sequences and series by looking at:

1. defining geometric sequences,
2. defining finite and infinite geometric series,
3. calculating finite geometric series,
4. calculating infinite geometric series,
5. looking at where this applies to engineering,
6. a summary, and
7. exercises.

1. Geometric sequences

A finite geometric sequence $x$ is any sequence of the form:

$x = (x[0], x[0]r, x[0]r^2, \ldots, x[0]r^n)$

where $x[0]$ and $r$ are real or complex numbers. We will describe the number $r$ as the ratio. Such a sequence is defined as the next entry being the ratio multiplied by the previous entry, so given $x[0]$, we define $x[k] = rx[k - 1]$. In this case, too, it is very straight-forward to come up with an explicit formula for such a sequence:

$x[k] = x[0]r^k$.

We may also consider, under certain circumstances, the infinite sequence

$x = (x[0], x[0]r, x[0]r^2, x[0]r^3, x[0] r^4, \ldots)$

2. Definitions of finite and infinite geometric series

Now, there are many situations in the engineering world where such sequences occur, and there are many times where we may want to compute the sum of such a progression:

$x[0] + x[0]r + x[0]r^2 + \cdots + x[0]r^n$

or the corresponding infinite sum

$x[0] + x[0]r + x[0]r^2 + x[0]r^3 + x[0] r^4 + \cdots$

In order to calculate such a series, we note that we may factor out the common factor of $x[0]$:

$x[0](1 + r + r^2 + \cdots + r^n)$

or

$x[0](1 + r + r^2 + r^3 + r^4 + \cdots)$.

Thus, we are really only interested in knowing the value of

$1 + r + r^2 + \cdots + r^n$

or if possible

$1 + r + r^2 + r^3 + r^4 + \cdots$.

This arithmetic series may also be represented by the sum

$\sum_{k = 0}^n r^k$ or $\sum_{k = 0}^\infty r^k$,

meaning that, in the first case, the variable $k$ takes on all values between $1$ to $n$, and we are then summing those values together; and in the second, the variable $k$ takes on all positive integers (including zero) and summing all these values together.

3. Calculating finite geometric series

We will begin by finding and proving a formula for the finite sum. We will then see how we can generalize this for the infinite sum, at least in certain circumstances. To find a formula for the finite sum, we will multiply the series by $1 = \frac{1 - r}{1 - r}$:

$\frac{1 - r}{1 - r}(1 + r + r^2 + r^3 + \cdots + r^n)$.

If we expand the numerator, we get

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

Multiplying through by $-r$, we get

$\frac{(1 - r)(1 + r + r^2 + r^3 + \cdots + r^n)}{1 - r} = \frac{(1 + r + r^2 + r^3 + \cdots + r^n) - r - r^2 - r^3 - r^4 - \cdots - r^{n + 1})}{1 - r}$.

Now, if you rearrange the numerator, you will see that it equals

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

You will see that all the terms cancel, so that we have the simpler formula

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

Thus, we have the formula

$\sum_{k = 0}^n r^k = \frac{1 - r^{n + 1}}{1 - r}$.

Thus, the more general geometric series has the formula

$\sum_{k = 0}^n x[0] r^k = x[0] \frac{1 - r^{n + 1}}{1 - r}$.

Proof by induction

We will now prove this formula is correct using induction:

Theorem
The geometric series $\sum_{k = 1}^n r^k = \frac{1 - r^{n + 1}}{1 - r}$.

Proof:
To prove a statement is true by induction, we must:

1. Prove that the statement is true for the base case, in this case, for $n = 1$.
2. Assume the statement is true for an arbitrary value of $n$, and
3. Under the assumption that it is true for $n$, we must show that the statement continues to be true for $n + 1$.

In this case, the statement for $n = 1$ is true:

$\sum_{k = 0}^0 r^k = r^0 = 1$,

and

$\frac{1 - r^{0 + 1}}{1 - r} = \frac{1 - r}{1 - r} = 1$.

As these are the same, the statement is correct when $n = 0$.

Next, we assume the statement is true for $n$, so then we consider the statement for $n + 1$: we want to show that

The arithmetic series $\sum_{k = 1}^{n + 1} k = \frac{1 - r^{(n + 1) + 1}}{1 - r} = \frac{1 - r^{n - 2}}{1 - r}$.

Now,

$\sum_{k = 1}^{n + 1} r^k = 1 + r + r^2 + r^3 + \cdots + r^{n - 1} + r^n + r^{n + 1}$,

so we see that

$\sum_{k = 1}^{n + 1} r^k = \left(\sum_{k = 1}^n r^k\right) + r^{n + 1}$.

By assumption, the sum up to $n$ is $\frac{1 - r^{n + 1}}{1 - r}$, so we can substitute this in:

$\sum_{k = 1}^{n + 1} r^k = \frac{1 - r^{n + 1}}{1 - r} + r^{n + 1}$,

so finding a common denominator, we have:

$\sum_{k = 1}^{n + 1} r^k = \frac{1 - r^{n + 1} + (1 - r)r^{n + 1}}{1 - r} = \frac{1 - r^{n + 1} + r^{n + 1} - r^{n + 2}}{1 - r}$.

The two intermediate terms cancel, so we are left with

$\sum_{k = 1}^{n + 1} r^k = \frac{1 - r^{n + 2}}{1 - r}$,

which is the statement evaluated at $n + 1$, so what we set out to prove. ▮

Using a proof by induction to show a statement is wrong

Let us consider a slightly different case: suppose someone comes to you with the claim:

$\sum_{k = 1}^n r^k = 1 + \frac{rn(rn - n - r + 3)}{2}$,

and that person demonstrates that it is indeed correct for $n = 1, 2$ and $3$. If you were to try to prove this by induction, we have the base case, so now we continue:

$\sum_{k = 1}^{n + 1} k = 1 + \frac{r(n + 1)(r(n + 1) - (n + 1) - r + 3)}{2} = 1 + r + \frac{rn(rn - n + r + 1)}{2}$.

Substituting in our assumed formula into the equation

$\sum_{k = 1}^{n + 1} r^k = \left(\sum_{k = 1}^n r^k\right) + r^{n + 1}$

yields

$\sum_{k = 1}^{n + 1} r^k = 1 + \frac{rn(rn - n - r + 3)}{2} + r^{n + 1}$.

This does not appear to be equal to the formula we expect, and therefore, let's subtract the two:

$\left(1 + r + \frac{rn(rn - n + r + 1)}{2}\right) - \left( 1 + \frac{rn(rn - n - r + 3)}{2} + r^{n + 1} \right)$.

When all else is cancelled, we are left with

$r(1 - r^n + n(r - 1))$

which is not identically zero. Thus, the two formulas are not equal, and thus we have a contradiction.

The only assumption we made was that $\sum_{k = 0}^n r^k = 1 + \frac{rn(rn - n - r + 3)}{2}$, and therefore this assumption must be wrong.

Therefore, $\sum_{k = 0}^n r^k \ne 1 + \frac{rn(rn - n - r + 3)}{2}$ and the other person is wrong. ▮

4. Calculating infinite geometric series

If $|r| < 1$, then as $n$ becomes large, $r^n$ (and so $r^{n + 1}$) tends to go to zero. Therefore, as $n$ becomes larger and larger, $\frac{1 - r^{n + 1}}{1 - r}$ comes closer and closer to $\frac{1}{1 - r}$.

We can show this by observing that

$\left(\sum_{k = 0}^n r^k \right) - \frac{1}{1 - r} = \frac{1 - r^{n + 1}}{1 - r} - \frac{1}{1 - r} = \frac{r^{n + 1}}{1 - r}$.

The denominator is fixed by our choice of $r$, so the only value that changes is the numerator, which we have observed becomes arbitrarily small as $n$ becomes larger. Thus, the difference between the finite sum and the value $\frac{1}{1 - r}$ becomes arbitrarily small.

5. Where does this apply in engineering?

When you play a game, or do anything, there is almost always an end-point in your mind. In engineering, however, you will be designing machines that are intended to work for arbitrary long periods of time. Consequently, you cannot assume that, for example, the pacemaker will be working for $365$ days.

The energy and length (or norm) of a signal

Given a discrete signal

$x = (x[0], x[1], x[2], x[3], x[4], \ldots)$,

we define the energy to be

$\sum_{k = 0}^\infty x[k]^2$

and the length or norm of the signal is given by

$||x|| = \sqrt{\sum_{k = 0}^\infty x[k]^2}$.

If you have already studied vectors, the Euclidean length of a two-dimensional vector $\vec{u} = \begin{pmatrix}u_1\\u_2\end{pmatrix}$ is $||\vec{u}|| = \sqrt{u_1^2 + u_2^2}$ and the length of a three-dimensional vector $\vec{v} = \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}$ is $||\vec{v}|| = \sqrt{v_1^2 + v_2^2 + v_3^2}$. Suppose that $\vec{u}$ gave the velocity of a car with the car moving $u_1 {\rm km/h}$ in the east-west direction (east being positive) and $u_2 {\rm km/h}$ in the north-south direction (north being positive). Thus, the speed of the car is $||\vec{u}|| = \sqrt{u_1^2 + u_2^2}$.

You will recall that the energy of a moving particle in one direction is $\frac{1}{2}mv^2$ where $m$ is the mass of the object and $v$ is the speed. Thus, the energy of a moving car is $\frac{1}{2}m_{\rm car}(u_1^2 + u_2^2)$.

Similarly, if $\vec{v}$ described the velocity of an aircraft in each of the three directions (including up-down with up being positive), then the speed is $||\vec{v}|| = \sqrt{v_1^2 + v_2^2 + v_3^2}$ gives the speed of the aircraft and the energy is $\frac{1}{2}m_{\rm aircraft}(v_1^2 + v_2^2 + v_3^2)$.

You will note that the energy of a discrete signal has a similar definition, only we sum the squares of all of the entries and the norm of the signal is the square root of this value.

In all real situations, such discrete signals are always finite in length; for example, you may have collected all such temperatures between January 17th, 1985

Suppose we have an infinite geometric signal

$x = (x[0], x[0]r, x[0]r^2, x[0]r^3, x[0]r^4, \ldots)$

where $|r| < 1$. In engineering, you will often want to calculate the norm and energy of such a signal, and the energy is

$\sum_{k = 0}^\infty \left( x[0]r^k \right )^2$,

while the norm is the square root of this sum-of-squares.

We can write this sum-of-squares as

$\sum_{k = 0}^\infty \left( x[0]r^k \right)^2 = \sum_{k = 0}^\infty x[0]^2 (r^k)^2$
$= x[0]^2 \sum_{k = 0}^\infty (r^2)^k = x[0]^2\frac{1}{1 - r^2}$

and the norm is therefore

$||x|| = \sqrt{x[0]^2 \frac{1}{1 - r^2}} = \left|x[0]\right|\sqrt{\frac{1}{1 - r^2}}$.

Thus, for example, the energy of the signal

$x = (0.70, -0.14, 0.028, -0.0056, 0.00112, -0.000224, \ldots)$

would be to observe that $x[0] = 0.7$ and $r = -0.2$, so $r^2 = 0.04$ and the energy is

$0.7^2\frac{1}{1 - (-0.2)^2} = 0.49\frac{1}{0.96} \approx 0.5104$

and the norm is the square root of this, or $||x|| \approx 0.7144$.

Similarly, with the signal

$y = (0.50, 0.35, 0.245, 0.1715, 0.12005, 0.084035, \ldots)$

we see that $y[0] = 0.5$ and $r = 0.7$, and so the energy is

$0.5^2\frac{1}{1 - (0.7)^2} = 0.25\frac{1}{0.51} \approx 0.4902$

and the norm is $||y|| \approx 0.7001$.

Similarly, with the signal

$z = (-0.30, 0.27, -0.243, 0.2187, -0.19683, 0.177147, \ldots)$

we see that $z[0] = -0.3$ and $r = -0.9$, and so the energy is

$(-0.3)^2\frac{1}{1 - (-0.9)^2} = 0.09\frac{1}{0.19} \approx 0.4737$

and the norm is $||z|| \approx 0.6882$.

Finally, with the signal

$w = (1.2, 0.012, 0.00012, 0.0000012, 0.000000012, \ldots)$

we see that $w[0] = 1.2$ and $r = 0.01$, and so the energy is

$1.2^2\frac{1}{1 - (0.01)^2} = 1.44\frac{1}{0.9999} \approx 1.440$

and the norm is $||w|| \approx 1.200$.

Dot product or inner product

Given the finite discrete signal

$x = (1.2, 0.9, 1.1, 1.2, 0.9, 0.8, 0.9, 1.1, 0.9, 0.9)$

and the two signals

$y = (1, 1, 1, 1, 1, 1, 1, 1, 1, 1)$
$z = (1, 0, 1, 0, 1, 0, 1, 0, 1, 0)$,

if you were asked which of $y$ and $z$ are more similar to $x$, you would likely say that $x$ is more similar to $y$. The question is, however, is there a way of measuring the similarity? For this, we define the inner product (sometimes called the dot product in secondary school).

Given two finite discrete signals

$x = (x[0], x[1], x[2], x[3], x[4], \ldots, x[n])$ and
$y = (y[0], y[1], y[2], y[3], y[4], \ldots, y[n])$,

we define the inner product (or "dot" product) to be

$\sum_{k = 0}^n x[k] y[k]$.

Similarly, given two infinite discrete signals

$x = (x[0], x[1], x[2], x[3], x[4], \ldots)$ and
$y = (y[0], y[1], y[2], y[3], y[4], \ldots)$,

we define the inner product (or "dot" product) to be

$\sum_{k = 0}^\infty x[k] y[k]$

but only if this infinite sum is finite (and not infinity).

Now, given two infinite geometric discrete signals:

$x = (x[0], x[0]r, x[0]r^2, x[0]r^3, x[0]r^4, \ldots)$ and
$y = (y[0], y[0]s, y[0]s^2, y[0]s^3, y[0]s^4, \ldots)$,

where $|r| < 1$ and $|s| < 1$. The inner product is therefore

$\sum_{k = 0}^\infty \left(x[0]r^k\right)\left(y[0]s^k\right) = x[0]y[0] \sum_{k = 0}^\infty (rs)^k = \frac{x[0]y[0]}{1 - rs}$.

Now, the inner product is equal to $0$ if and only if $x[0] = 0$ or $y[0] = 0$. Note also that if $|r| < 1$ and $|s| < 1$, then $|rs| = |r||s| < 1$. The inner product is therefore always a non-zero real number (remembering that $\pm \infty$ is not a real number).

For example, given the two signals

$x = (0.70, -0.18, 0.036, -0.0072, 0.00144, -0.000288, \ldots)$ and
$y = (0.50, 0.35, 0.245, 0.1715, 0.12005, 0.084035, \ldots)$,

would be to observe that $x[0] = 0.9$, $r = -0.2$, $y[0] = 0.5$ and $s = 0.7$, and so the inner product is

$(0.9)(0.5)\sqrt{ \frac{1}{1 - (-0.2)(0.7)} } = 0.45\sqrt{\frac{1}{1.14} } \approx 0.3947$.

The delay operator

Given an arbitrary signal

$x = (x[0], x[1], x[2], x[3], x[4], \ldots)$,

we delay the signal by dropping the first entry:

$Dx = (x[1], x[2], x[3], x[4], x[5], \ldots)$.

Now, if we represent the entire signal $(x[0], x[1], x[2], x[3], x[4], \ldots)$ by the letter $x$, then we note that we can represent

$x = (x[0], x[1], x[2], x[3], x[4], \ldots)$

by the letter $x$, then

$Da = (x[1], x[2], x[3], x[4], x[5], \ldots)$.

Now, if $x$ is geometric in nature, then

$Dx = (x[0]r, x[0r^2, x[0]r^3, x[0]r^4, x[0]r^5, \ldots) = r(x[0], x[0]r, x[0]r^2, x[0]r^3, x[0]r^4, \ldots) = rx$

where $rx$ represents multiplying each entry in the sequence $x$ by the scalar value $r$. That is, delaying the signal $x$ is equivalent to multiplying each entry by $r$.

Note that the inner product of $x$ and $Da$ is

$\sum_{k = 0}^\infty \left(x[0]r^k\right)\left(x[0]r^{k + 1}\right) = x[0]^2r \sum_{k = 0}^\infty (r^2)^k = \frac{x[0]^2 r}{1 - r^2}$,

the sign of which depends only on $r$.

6. Summary

Finite geometric sequences are of the form

$(a[0], a[0]r, a[0]r^2, a[0]r^3, \ldots, a[0]r^n$

while infinite geometric sequences are of the form

$(a[0], a[0]r, a[0]r^2, a[0]r^3, a[0]r^4, \ldots$.

If the ratio $|r| \le 1$, then the infinite geometric series is bounded, and if the ratio $|r| > 1$, then the infinite geometric series is unbounded.

The sum of a finite geometric sequences is defined as a finite geometric series and can be calculated as

$\sum_{k = 0}^n a[0]r^k = a[0]\frac{1 - r^{n + 1}}{1 - r}$,

while the sum of an infinite geometric sequences when the ratio $r$ satisfies $|r| < 1$ is defined as a infinite geometric series and can be calculated as

$\sum_{k = 0}^\infty a[0]r^k = a[0]\frac{1}{1 - r}$.

7. Exercises

1. Calculate each of the following series:

• $1 + 0.2 + 0.04 + 0.008 + 0.0016 + 0.00032$
• $1 - 0.3 + 0.09 + 0.027 + 0.0081$
• $1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512$
• $1 - 10 + 100 - 1000 + 10000 - 100000 + 1000000 - 10000000$

2. Notice that $10^3 = 1000$ and $2^10 = 1024$, so $2^10 \approx 10^3$. Thus, $2^20$ is approximately one million, $2^30$ is approximately one billion, and $2^40$ is approximately one trillion. Thus, what is the series

$1 + 2 + 4 + \cdots + 2^31$?

3. What are the finite geometric series:

• $3.2 + 6.4 + 12.8 + 25.6 + 51.2 + 102.4 + 204.8$
• $-0.63 + 1.26 - 2.52 + 5.04 - 10.08 + 20.16 - 40.32 + 80.64 - 161.28$
• $95.0 + 19.0 + 3.80 + 0.760 + 0.1520 + 0.03040 + 0.006080 + 0.0012160 + 0.00024320$
• $54.0 - 21.6 + 8.64 - 3.456 + 1.3824 - 0.55296$

4. What is the infinite geometric series

• $1 + 0.5 + 0.25 + 0.125 + 0.0625 + 0.03125 + 0.015625 + \cdots$
• $1 - 0.5 + 0.25 - 0.125 + 0.0625 - 0.03125 + 0.015625 - \cdots$

5. Of the four series in Question 3, for which can the corresponding infinite geometric series be calculated, and what is the value of that series?

6. Show that the formula for a finite geometric series for correct when we have

$(-1)^0 + (-1)^1 + (-1)^2 + (-1)^3 + \cdots (-1)^n = 1 - 1 + 1 - 1 + \cdots + (-1)^n$.

7. The formula for the geometric series fails when $r = 1$. Why does this fail, and what is the value of the geometric series when $r = 1$.