Lesson

An ordered list of numbers, separated by commas, is called a progression or sequence. For example $-3,5,13,21,29$−3,5,13,21,29 and $1,10,100,1000...$1,10,100,1000... are two interesting mathematical progressions.

If the sequence ends, such as the first sequence above, it is known as a finite sequence. Otherwise, it is said to be infinite.

You might notice that in both examples there is a pattern in the way the progressions are formed. When a pattern is detectable in a progression, a generating rule can often be established that allows us to determine any term in the sequence.

Mathematicians can sometimes develop explicit formulas that allow the calculation of any particular term in the sequence.

For example the rule $a_n=\sqrt{n}$`a``n`=√`n` means that the $n$`n`th term is the square root of n. So the first term becomes $a_1=\sqrt{1}=1$`a`1=√1=1 and the second term $a_2=\sqrt{2}$`a`2=√2 etc., so that the sequence becomes $1,\sqrt{2},\sqrt{3},2,...$1,√2,√3,2,...and so on.

We can express a sequence using a recursive formula or a recurrence relation where each new term is generated by some function of a previous term or terms.

Take, for example, the sequence described by:

$a_n=2a_{n-1}+n,a_1=3$`a``n`=2`a``n`−1+`n`,`a`1=3

This expression tells us that the first term $a_1$`a`1 is $3$3. Then it shows us how to find, progressively, any other term in the sequence. For example, looking at the formula, the second term $a_2$`a`2 is equal to twice the first term $a_1$`a`1 plus $2$2, which is $2\times3+2$2×3+2 or $8$8. The third term is equal to twice the second term plus $3$3, or in other words $19$19. The fourth term is twice $19$19 plus $4$4, or $42$42. The fifth term is twice $42$42 plus $5$5, or $89$89. This process of deducing the $n$`n`th term from the $\left(n-1\right)$(`n`−1)th term can continue indefinitely. A simple computer spreadsheet program could develop a list of terms very easily.

Is the sequence $1,2,3,4,5,6$1,2,3,4,5,6 finite or infinite?

Finite

AInfinite

BFinite

AInfinite

B

State the first five terms of the sequence $a_n=3n-3$`a``n`=3`n`−3.

Write all five terms on the same line separated by a comma.

Consider the sequence defined by $a_1=3$`a`1=3, $a_n=a_{n-1}+n$`a``n`=`a``n`−1+`n` for $n>1$`n`>1.

State the first term.

State the second term.

State the third term.

State the fourth term.

A series is the indicated sum of all of the terms of a sequence. Like sequences, series can be finite or infinite. A finite series is the indicated sum of all the terms of a finite sequence, and an infinite series is the indicated sum of all the terms of an infinite sequence.

The sum of the first $n$`n` terms of a series is called the $n$`n`th partial sum and is denoted $S_n$`S``n`. The $n$`n`th partial sum of any series can be found by calculating each term up to the $n$`n`th term and then finding the sum of those terms.

Writing out the partial sum of a sequence can take up a lot of space and time. A number of centuries ago, mathematicians succeeded in developing a shorthand notation for many sequences, and they borrowed from the Greek alphabet to do it.

The symbol $\Sigma$Σ (pronounced "sigma") is the capital letter S in the Greek alphabet. When $\Sigma$Σ is used to express a series in mathematics, then in those instances it stands for the word "Sum". To explain how the symbol is used, consider the following expression:

$\sum_{n=1}^{n=5}$`n`=5∑`n`=1 $n^2$`n`2

The equation $n=1$`n`=1 directly below and to the right of the $\Sigma$Σ sign tells us that we start the series by substituting $n=1$`n`=1 into the sequence formula shown as $n^2$`n`2. So the series begins:

$1^2$12

Then we increase $n$`n` by $1$1 so that $n$`n` becomes $2$2. The new value of $n$`n` is substituted into the sequence formula to reveal $2^2$22 so that the series becomes:

$1^2+2^2$12+22

Again we increase $n$`n` by $1$1 and substitute for the third term so that the series becomes:

$1^2+2^2+3^2$12+22+32

This process of increasing $n$`n` by $1$1 continues until $n=5$`n`=5 is reached as shown directly above and to the right of the $\Sigma$Σ sign. The complete sum becomes:

$\sum_{n=1}^{n=5}n^2=1^2+2^2+3^2+4^2+5^2$`n`=5∑`n`=1`n`2=12+22+32+42+52

Thus this series sums to $55$55.

When it is clear that the variable being referred to is $n$`n`, we are allowed to drop the "$n$`n`" in the superscripted and subscripted expression so that the above example can be written:

$\sum_1^5n^2=1^2+2^2+3^2+4^2+5^2$5∑1`n`2=12+22+32+42+52

As a second example, the first $100$100 terms of the arithmetic series whose first term is $a=10$`a`=10 and whose common difference is $d=3$`d`=3 is given, with our new notation, as:

$\sum_1^{100}\left(7+3n\right)$100∑1(7+3`n`)

Following the same strategy, you should be able to see that the series written out would look like:

$10+13+16+...+304+307$10+13+16+...+304+307

Note that the common difference, the first term and last term are easy to spot, and this means we can use the sum formula for an arithmetic progression to show that:

$S_{100}=\frac{100}{2}\left(10+307\right)=15850$`S`100=1002(10+307)=15850

In some instances of geometric series, we can add up an infinite number of terms and obtain a finite sum. Specifically this happens when the common ratio $r$`r` is between, but not including, $-1$−1 and $1$1.

Take for example the expression:

$\sum_{n=1}^{\infty}32\times\left(\frac{1}{2}\right)^{n-1}$∞∑`n`=132×(12)`n`−1

Using our strategy, we can write the first few terms as $32+16+8+4+...$32+16+8+4+... and using the limiting sum formula, we find that the finite sum is given by:

$S_{\infty}=\frac{a}{1-r}=\frac{32}{1-\left(\frac{1}{2}\right)}=64$`S`∞=`a`1−`r`=321−(12)=64

As a final note, the summation notation is quite versatile in its application to series. For example, the mean of a set of scores, say in general terms the scores $x_1,x_2,x_3,...,x_n$`x`1,`x`2,`x`3,...,`x``n` is their sum divided by $n$`n`.

We can write this using what is known as a dummy variable $i$`i` and write:

$\sum_{i=1}^n$`n`∑`i`=1 $\frac{x_i}{n}$`x``i``n`

In expanded form this expression reads:

$\frac{x_1}{n}+\frac{x_2}{n}+\frac{x_3}{n}+...+\frac{x_n}{n}$`x`1`n`+`x`2`n`+`x`3`n`+...+`x``n``n`

or equivalently $\frac{x_1+x_2+x_3+...+x_n}{n}$`x`1+`x`2+`x`3+...+`x``n``n`.

Write the following series using sigma notation.

$4+8+12+16+20+\text{. . .}$4+8+12+16+20+. . .

$\sum_{k=1}^{\infty}\left(\editable{}\right)$∞∑

`k`=1()

Consider the series:

$\frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\frac{1}{5^3}$113+123+133+143+153

Rewrite the series using sigma notation in the form $\sum_{k=\editable{}}^{\editable{}}\editable{}$∑`k`=.

Find the value of $\sum_{r=1}^4\frac{1}{r+2}$4∑`r`=11`r`+2.