# Sequences and Series

## Sequences

A sequence $\{ a_{n} \}$ is an infinite list of numbers $$a_{1}, a_{2}, a_{3}, \ldots,$$ where we have one number $a_{n}$ for every positive integer $n$.

### Defining sequences.

We can specify a sequence in various ways.

Pattern. We can specify it by listing some elements and implying that the pattern shown continues.

### Example.

For example $$2, 4, 6, 8, \ldots$$ would be the sequence consisting of the even positive integers.

Formula. We can also specify a sequence by giving a formula for the term that corresponds to the integer $n$.

### Example.

For example the sequence $$2, 4, 6, 8, \ldots$$ can also be specified by the explicit formula $$a_{n} = 2n.$$ Recursively. Finally, we can also provide a rule for producing the next term of a sequence from the previous ones. This is called a recursively defined sequence.

### Example.

For example the sequence $$2, 4, 6, 8, \ldots$$ can be specified by the rule $$a_{1} = 2 \quad \text{ and } \quad a_{n} = a_{n-1} +2 \text{ for } n\geq 2.$$ This rule says that we get the next term by taking the previous term and adding $2$. Since we start at the number 2 we get all the even positive integers.

Let's discuss these ways of defining sequences in more detail, and take a look at some examples.

### Part 1: Arithmetic Sequences

The sequence we saw in the previous paragraph is an example of what's called an arithmetic sequence: each term is obtained by adding a fixed number to the previous term.

Alternatively, the difference between consecutive terms is always the same.

### General Formula.

If a sequence $a_{n}$ is arithmetic, then there is a fixed number $d$ so that $a_{n+1} -a_{n} =d$ for any $n.$ The number $d$ is usually called the step or difference. Let's try to find a formula for the term $a_{n}$ of an arithmetic sequence in terms of $d$ and $a_{1}$.

Let's start with $a_{n} = a_{n-1} + d$. Applying this again, we see that since $a_{n-1} = a_{n-2} + d$ so we get that $a_{n} = a_{n-2} + d +d = a_{n-2} + 2d$. We can continue this way and get: \begin{align*} a_{n} &= a_{n} = a_{n-1} + d \\ &= a_{n-2} + d +d = a_{n-2} + 2d \\ &= a_{n-3} + d + d = a_{n-3} + 3d \\ &\vdots \\ &= a_{2} + (n-2)d \\ &= a_{1} + (n-1)d \\ \end{align*} So we get that in an arithmetic sequence $a_{n}$ with steps of size $d$, the formula for $a_{n}$ is given by: $$a_{n} = a_{1} + (n-1)d$$

### Example. Consider the sequence $3, 8, 13, 18, 23, 28, \ldots$. Is it arithmetic? If so, find a formula for $a_{n}$, and use it to find $a_{101}$, the 101st term in the sequence.

Solution. This sequence is arithmetic, since the difference between each term is $5.$
($8-3 = 13-8 = 18-13 = \cdots = 5$.) So this is an arithmetic sequence with step $d=5$ and first term $a_{1} = 3$.
Our formula above gives $a_{n} = a_{1} + (n-1)d = 3 + (n-1)5$.
For $a_{101}$ we plug in $n=101$ into this formula to obtain $a_{101} = 3 + (100)5 = 503$.

### Part 2: Geometric Sequences

Consider the sequence $2, 4, 8, 16, 32, 64, \ldots$. This sequence is not arithmetic, since the difference between terms is not always the same. If we look closely, we will see that we obtain the next term in the sequence by multiplying the previous term by the same number. Equivalently, the ratio of consecutive terms is always the same (namely $2$).

A sequence $a_{n}$ where there is a fixed $r$ so that $\frac{a_{n}}{a_{n-1}} = r$ for all $n$ is called a geometric sequence. The number $r$ is usually called the ratio.

### General Formula.

Let's try to find the formula for the term $a_{n}$ of a geometric sequence in terms of $r$ and the first term.

Let's start with the relation $\frac{a_{n}}{a_{n-1}} = r$. This gives $a_{n} = r a_{n-1}$. Using this again, we get $a_{n} = r (ra_{n-2}) = r^{2} a_{n-2}$. We can continuer this way and get: \begin{align*} a_{n} &= r a_{n-1} \\ &= r^{2} a_{n-2} \\ &= r^{3} a_{n-3} \\ &\vdots \\ &= r^{n-2} a_{2} \\ &= r^{n-1} a_{1} \\ \end{align*} So we get that for a geometric sequence $a_{n}$ with ratio $r$, the formula for $a_{n}$ is given by: $$a_{n} =r^{n-1} a_{1}$$

### Example. Consider the geometric sequence $3, 6, 12, 24, 48, \ldots$. Find a formula for $a_{n}$ and use it to find $a_{7}$.

Solution. To find $r$, we should look at the ratio between successive terms: $r= \frac{a_{1}}{a_{2}} =\frac{6}{3} = 2$. Then using the formula above we get $a_{n} = r^{n-1} a_{1} = 2^{n-1} \cdot 3$.

To find $a_{7}$ we set $n=7$ and get $a_{7} = 2^{7-1} \cdot 3 = 2^{6} \cdot 3 = 64 \cdot 3 = 192$.

### Part 3: Recursive Sequences

We have already briefly discussed this idea in the first paragraph. We shall now discuss this in more detail, together with some extra examples.
As we saw in the section on geometric sequences, we can define a geometric sequence either by the rule $a_{n} = r^{n-1} a$, or by the rule that $a_{n} = r a_{n-1}$.
The latter rule is an example of a recursive rule. A recursively defined sequence, is one where the rule for producing the next term in the sequence is written down explicitly in terms of the previous terms.

Let's consider the following (rather famous) example.

### Example.

Define a sequence $a_{n}$ as follows: Let $$a_{1} = 1 \quad , \quad a_{2} = 1 \quad \text{, and} \quad a_{n} = a_{n-1} + a_{n-2} \text{ for } n\geq 2.$$ This rule says that to get the next term in the sequence, you should add the previous two terms. Since this rule requires two previous terms, we need to specify the first two terms of the sequence $a_{1}, a_{2}$ to get us started. Using this we can start to list the terms in the sequence, and get $1, 1, 2, 3, 5, 8, 13, 21, 34,\ldots$. (This is the well known Fibonacci sequence.)

### Example.

Consider the recursively defined sequence $$a_{1} = 1 \quad , \quad a_{2} = 1 \quad , \quad a_{3} = 1 \quad \text{, and} \quad a_{n} = \frac{a_{n-3}}{a_{n-1} + a_{n-2}} \text{ for } n\geq 3.$$ List the first 7 terms of this sequence.

Solution. \begin{align*} a_{1} &= 1 \\ a_{2} &= 1 \\ a_{3} &= 1 \\ a_{4} &= \frac{1}{1+1} = \frac{1}{2} \\ a_{5} &=\frac{1}{1+\frac{1}{2}} = \frac{1}{\frac{3}{2}} = \frac{2}{3} \\ a_{6} &= \frac{1}{\frac{1}{2} + \frac{2}{3}} = \frac{1}{\frac{7}{6}} = \frac{6}{7} \\ a_{7} &= \frac{ \frac{1}{2}}{\frac{2}{3} + \frac{6}{7}} = \frac{ \frac{1}{2}}{\frac{32}{21}} = \frac{21}{64} \end{align*}

### Part 4: Sequences via Lists

The method of using a list to specify a sequence perhaps is the most tricky, since it requires us to look at a short piece of a sequence, and guess at the pattern or rule that is being used to produce the terms in the sequence.
Now that we have seen some more examples of sequences we can discuss how to look for patterns and figure out given a list, how to find the sequence in question.

### Example.

When given a list, such as $1, 3, 9, 27, 81, \ldots$ we can try to look for a pattern in a few ways.
Now that we have seen arithmetic, geometric and recursive sequences, one thing we can do is try to check if the given sequence is one of these types.

Arithmetic? To check if a sequence is arithmetic, we check whether or not the difference of consecutive terms is always the same. In this case, the difference changes: $$a_2- a_1 = 3-1 =2 \neq 6 = 9-3 = a_3-a_2.$$ Geometric? To check if a sequence is geometric we check whether or not the ratio of consecutive terms is always the same. In the case it is, so we conclude that the sequence is geometric: $$\frac{3}{1}= \frac{9}{3} = \frac{27}{9} = \frac{81}{27} = 3.$$ This tells us that the sequence is geometric with ratio 3, and initial term 1, so we get that the sequence is given by $$a_{n} = 3^{n-1}.$$ This sequence can also be defined recursively, by the formula $$a_{1} =1 \quad \text{, and} \quad a_{n} = 3a_{n-1} \text{ for } n\geq 2.$$

### Example.

Consider the sequence $1, -3, -7, -11, -15, -19, -23, \ldots$. Determine a formula for the $n^{\text{th}}$ term in the sequence.

Solution. We quickly see that this series is not geometric, since $\frac{1}{-3} \neq \frac{-3}{-7}$.

We can now try to see if the sequence is arithmetic. If we look at the differences of consecutive terms, we get: $-3 - 1 = -4 = -7 - (-3) = -11 - (-7)$, so we see that this is an arithmetic sequence with difference $d=-4$. So the general term is $$a_{n} = a_{1} + (n-1) d = 1 +(n-1) (-4) = - 4n + 5.$$
(We could also try to identify a recursive definition of this sequence.)