This section covers:
 Introduction to Sequences and Series
 Summary of Formulas for Sequences and Series
 Sequences and Series Terms
 Explicit Formulas Versus Recursive Formulas
 Arithmetic Sequences
 Geometric Sequences
 Writing Formulas
 Arithmetic Series
 Summation Notation
 Geometric Series
 Applications of Sequences and Series
 More Practice
Introduction to Sequences and Series
Sequences and Series are basically just numbers or expressions in a row that make up some sort of a pattern; for example, January, February, March, …, December is a sequence that represents the months of a year. Each of these numbers or expressions are called terms or elements of the sequence.
Sequences are the list of these items, separated by commas, and series are the sum of the terms of a sequence (if that sum makes sense; it wouldn’t make sense for months of the year).
You may have heard the term inductive reasoning, which is reasoning based on patterns, say from a sequence (as opposed to deductive reasoning, which is reasoning from rules or definitions). For example, let’s try to find the next few terms in the following sequences:
Sequence  Next Two Terms 
1, 8, 27, __ , __  Every term is cubed. Next two terms are \({{4}^{3}},\,\,{{5}^{3}}\), or 64, 125. 
3, 7, 11, 15, __ , __  Every term is 4 more than the previous one. Just add 4 to get to the next term. Next two terms are \(15+4,\,\,15+4+4\), or 19, 23. 
1000, 100, 10, 1, ___ , ___  Every term is the previous term divided by 10, or .1 times the previous one. Just multiply by .1 to get to the next term. Next two terms are \(1\times .1,\,\,1\times .1\times .1\), or .1, .01. 
\(\displaystyle \frac{{{{x}^{3}}}}{{x+1}},\,\,\frac{{{{x}^{5}}}}{{{{{\left( {x+1} \right)}}^{2}}}},\,\frac{{{{x}^{7}}}}{{{{{\left( {x+1} \right)}}^{3}}}},\,\,\_\_\_\,,\,\_\_\_\)  Sometimes we need to look at both the top and bottom of a fraction, and different things are going on. On the top (numerator), the next term is the previous multiplied by \({{x}^{2}}\), and the bottom (denominator), the next term is the previous term multiplied by \(\left( {x+1} \right)\). The next two terms are \(\displaystyle \frac{{{{x}^{9}}}}{{{{{\left( {x+1} \right)}}^{4}}}},\,\,\frac{{{{x}^{{11}}}}}{{{{{\left( {x+1} \right)}}^{5}}}}\). 
\(\displaystyle \frac{1}{2},\,\frac{2}{4},\,\frac{3}{8},\,\,\_\_\,,\,\,\_\_\)  On the top (numerator), the next term is 1 more than the previous one, and the bottom (denominator), the next term is the previous term multiplied by 2. The next two terms are \(\displaystyle \frac{4}{{16}},\,\,\frac{5}{{32}}\). 
We will do more like this in the Writing Formulas section below.
Summary of Formulas for Sequences and Series
Before we get started, here is a summary of the main formulas for Sequences and Series:
\(\begin{array}{l}{{a}_{1}}=\,\,\text{first term}\\{{a}_{n}}=\,\,n\text{th term}\end{array}\)  Arithmetic d = common difference  Geometric r = common ratio 
Sequences 
\({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d\)
(Recursive: \({{a}_{n}}=a,\,\,\,\,\,{{a}_{n}}={{a}_{{n1}}}+d\))
 \({{a}_{n}}={{a}_{1}}{{\left( r \right)}^{{n1}}}\)
(Recursive: \({{a}_{n}}=a,\,\,\,\,\,{{a}_{n}}=r{{a}_{{n1}}}\) ) 
Finite Series (Sum) 
\(\displaystyle {{S}_{n}}=\frac{n}{2}\left( {{{a}_{1}}+{{a}_{n}}} \right)\,\,\,=\,\,\,\frac{n}{2}\left( {2{{a}_{1}}+\left( {n1} \right)d} \right)\)
 \(\displaystyle {{S}_{n}}=\frac{{{{a}_{1}}\left( {1{{r}^{n}}} \right)}}{{1r}}\) 
Infinite Series (Sum)  Not Applicable 
\(\displaystyle {{S}_{\infty }}=\frac{{{{a}_{1}}}}{{1r}},\,\,\,\,\,\,\,\left r \right<1\)

Sequences and Series Terms
OK, so I have to admit that this is sort of a play on words since each element in a sequence is called a term, and we’ll talk about the terms (meaning words) that are used with sequences and series, and the notation.
Let’s first compare sequences to relations or functions from the Algebraic Functions section. Think of the x part of the relation (the independent variable) as the numbers in a row that represent each part of the sequence; these typically start at 1 (one). So the x part will typically be 1, 2, 3, and so on. This is usually represented by the variable n.
Note that the n part will always be a positive integer.
Then think of the y part of the sequence as the actual term, or what the element is for that particular n value; this is usually represented by \({{a}_{n}}\) (but sometimes you see this as \(f\left( n \right)\) instead) .
So for the sequence 3, 7, 11, 15, and so one, we have:
So think of the “points” or “coordinates” of a sequence as \(\left( {n,\,\,{{a}_{n}}} \right)\).
Explicit Formulas versus Recursive Formulas
Explicit formulas are formulas that are computed for each term; in other words, you can look at the formula for a term and know exactly how to get that term (you don’t rely on the previous term). We’ll talk about the two main types of explicit formulas, Arithmetic and Geometric, later.
Recursive formulas are formulas that use the previous term to get to the next one, and we always have to indicate what the first term is (\({{a}_{{1}}}\)), so we can get started. Notice that to reference the previous term, we use \({{a}_{{n1}}}\) (sometimes \(f\left( {n1} \right)\)), which makes sense, since we need the term before this one.
Since the use of recursive formulas are very limited, we typically don’t deal with them very much.
Here are examples; for most of these, we’ll come up with some tools to make finding the formulas much easier. And don’t worry if you don’t see how to get these right away; these take practice!
Arithmetic Sequences
Arithmetic sequences are those where the difference between the terms is always the same. They can be defined recursively as \({{a}_{1}}=\,\,a;\,\,\,\,{{a}_{n}}={{a}_{{n1}}}+\,\,d,\,\,\text{for}\,\,\,n>1\). Here, “a” is the first term, and “d” is the common difference.
The way I like to tell if a sequence is arithmetic is to see if “second term – (minus) first term” equals “third term – second term” equals “fourth term – third term”, and so on. Then once we have this number that is always the same, we have the common difference, or “d”. Note that “d” can be negative, a fraction, or a decimal. (“n” has to always be a positive integer).
Formula for an Arithmetic Sequence
We saw above that the if difference between the terms is d for an arithmetic sequence, the explicit formula then will have a “dn” in it. We won’t prove it here, but it turns out the explicit formula for an arithmetic sequence is the formula below, for each term \({{a}_{n}}\). This makes sense since we always start out with the first term, and then we are adding the common difference d “n – 1” more times!
This is called the formula for a general term for an arithmetic sequence, and you’ll want to memorize this formula.
Now let’s do some problems:
It turns out that if we are given the value of two specific terms of a sequence (and what terms they are – the “n”), we can derive the equation of that sequence. What we can do is subtract the n’s and subtract the terms (\({{a}_{n}}\))’s and then divide these two numbers to get d, the common difference. This makes sense since we are multiplying by d to get the successive terms.
Then to get the first term (\({{a}_{1}}\)), we can use the formula \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d\), using one of the \({{a}_{1}}\) and n combinations, and the d we just got. Pretty cool!
So think of the common difference d as a slope of this “linear equation”; we are finding this slope using our familiar slope formula for two points of the form \(\left( {n,\,\,{{a}_{n}}} \right)\) as change of y’s over change of x’s : \(\displaystyle d=\,\frac{{{{a}_{n}}\,\text{(second)}{{a}_{n}}\text{(first)}}}{{n\text{(second)}n\text{(first)}}}\)
Let’s try some problems:
Here’s one more type of problem you may see:
Geometric Sequences
Whereas arithmetic sequences are those where the difference between the terms is the same, geometric sequences are those where the quotient of the terms is always the same. They can be defined recursively as \({{a}_{1}}=\,\,a;\,\,\,\,{{a}_{n}}=r{{a}_{{n1}}},\,\,\text{for}\,\,\,n>1\). Here, “a” is the first term, and “r” is the common ratio. Note that “r” can be negative, a fraction, or a decimal. (“n” has to always be a positive integer).
The way I like to tell if a sequence is geometric is to see if “second term /(divided by) first term” equals “third term/second term” equals “fourth term/third term”, and so on. Then once we have this number that is always the same, we have the common ratio!
Formula for a Geometric Sequence
It turns out the explicit formula for a geometric sequence is the formula below, for each term . This makes sense since we always start out with the first term, and then we are multiplying the common ratio r “n – 1” more times!
This is called the formula for a general term for a geometric sequence, and you’ll want to memorize this formula.
So what this says is the nth term of an arithmetic sequence is the first time times r (common ratio) raised to the (n – 1).
Now let’s do some problems:
Just like for arithmetic sequences, it turns out that if we are given the value of two specific terms of a sequence (and what terms they are – the “n”), we can derive the equation of that sequence. What we can do is subtract the n’s (say get x) and divide the terms (\({{a}_{n}}\) )’s (say get y) and then take the x’th root of y to get r, the common ratio (and we have to watch even roots, as shown in an example below). This makes sense since we are raising r to successive powers to get the successive terms.
Then to get the first term (\({{a}_{1}}\) ), we can use the formula , using one of the \({{a}_{1}}\) and n combinations, and the r we just got. Pretty cool!
Let’s try some problems:
Here’s one more type of problem you may see:
Writing Formulas
Now let’s try to distinguish between arithmetic and geometric sequences, and also write formulas for those sequences that are neither arithmetic nor geometric.
Remember again to look for second – first = third – second, and so on, for arithmetic sequences, and second / first = third / second, and so on for arithmetic sequences.
If neither of these work, look for squares or cubes in a row, added to or subtracted by a certain number.
Here’s another trick, if it appears the sequence is a quadratic. Let’s use this example: 3, 6, 10, 15 ,21, …
Of course, it might easier to just do a regression on your calculator with the sequence: use the n for x values and \({{a}_{n}}\) for y values. We did regressions like this here in the Scatterplots, Correlation, and Regression section.
Here are some problems; we’ll be working again with explicit formulas (as opposed to recursive):
Arithmetic Series
When we add up some or all of the terms of a sequence, we have a series. So when we add up the terms of an Arithmetic Sequence, we have an Arithmetic Series.
For example, the formula for the arithmetic sequence with \({{a}_{1}}=1\) and \(d=2\) is \({{a}_{n}}=1+\left( {n1} \right)2=2n1\). For n =1 to 5, this sequence is 1, 3, 5, 7, 9. We would write the arithmetic series as the sum of these elements, or 1 + 3 + 5 + 7 + 9. The partial sum (since we are only adding up 5 terms) of this series is 1 + 3 + 5 + 7 + 9 = 25.
Another way to write this is to use summation notation (also called the summation formula) \(\sum\limits_{{k=1}}^{5}{{\,1+\left( {k1} \right)2}}=\sum\limits_{{k=1}}^{5}{{\,2k1}}\) (we usually use the variable “k”). The sigma sign, or \(\sum{{}}\) symbol means take the equation after it, start with the lower bound, or what’s at the bottom of the sigma sign (1 in this case), plug in for n, and add up everything until you get to the upper bound, or what’s at the top of the sigma sign (5 in this case).
Here’s an example of how we might use an arithmetic series. Let’s say we have a theater that we are building and we know every row has 2 more seats than the row before it and there are 20 seats in the first row. We also know that we want 100 rows of the theater, and we are about to order the total number of seats in the theater. Do you see how we would then need to get the sum of an arithmetic series of 100 terms (with common difference 2) to find out how many seats to order? This series would be \(\sum\limits_{{k=1}}^{{100}}{{\,20+\left( {k1} \right)}}2\).
Formula for an Arithmetic Series
We actually have a formula to get this sum; this is called a partial sum formula. Note that we can never get the total or infinite sum of an arithmetic series, since this sum would just go on forever (extremely large or extremely small). We’ll talk how this will be a limit later.
Here is the partial sum formula for an arithmetic sequence, and you’ll want to memorize this formula:
\(\displaystyle {{S}_{n}}\,\,=\,\,\frac{n}{2}\left( {{{a}_{1}}+{{a}_{n}}} \right),\,\,\,\,\text{where }{{a}_{n}}\,\,=\,\,{{a}_{1}}+\left( {n1} \right)d\) or \(\displaystyle {{S}_{n}}\,\,=\,\,\frac{n}{2}\left[ {2{{a}_{1}}+\left( {n1} \right)d} \right]\)
Notice how the we’re using the formula for the n^{th} term of an arithmetic sequence (\({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d\)) to get the second formula. (Personally, I prefer not to use the second formula, but just memorize the first, since we already have memorized the \({{a}_{n}}\) formula.)
It’s pretty straightforward to use this formula to solve partial sum arithmetic series problems:
Problem  Solution 
Find an equation both for the nth term of the arithmetic sequence and the nth partial sum \({{S}_{n}}\) with:
\({{a}_{1}}=2,\,\,\,\,d=4\)
Then find \({{S}_{{40}}}\).  We know the formula for the nth term for an arithmetic sequence is \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d\), so for this sequence we have \({{a}_{n}}=2+\left( {n1} \right)4;\,\,{{a}_{n}}=4n2\). The sequence looks like 2, 6, 10, …, and the series looks like 2 + 6 + 10 + …., which is also \(\displaystyle \sum\limits_{{k=1}}^{n}{{2+\left( {k1} \right)4}}=\sum\limits_{{k=1}}^{n}{{4k2}}\). The equation then for the series’ nth partial sum is \({{S}_{n}}\,\,=\,\,\frac{n}{2}\left[ {2+\left( {4n2} \right)} \right]=\frac{n}{2}\left( {4n} \right)=2{{n}^{2}}\). To get the sum of the first 40 terms, we have \({{S}_{{40}}}\,=\,2{{\left( {40} \right)}^{2}}=3200\). 
Find an equation for the nth partial sum \({{S}_{n}}\):
\(100+75+50+…\)  We can see that this is an arithmetic series, since the second term – the first = the third term – the second = –25; this is d, or the common difference. The first term is 100, so \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d=100+\left( {n1} \right)\left( {25} \right)=25n+125\). The equation for this series’ partial sum is \({{S}_{n}}\,\,=\,\,\frac{n}{2}\left[ {100+\left( {25n+125} \right)} \right]=\frac{n}{2}\left( {25n+225} \right)\). (We can also write this \(\displaystyle {{S}_{n}}\,\,=\,\frac{{25{{n}^{2}}+225n}}{2}\).) Try this for n = 2 (\({{S}_{n}}=175\)); it works! 
Find an equation for the nth partial sum \({{S}_{n}}\), and find \({{S}_{{20}}}\):
\(\left( {5+t} \right)+\,\,\left( {5+3t} \right)+\,\,\left( {5+5t} \right)+…\)  We can see that this is an arithmetic sequence with d = 2t. The first term is “5 + t”, so we have \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d=\left( {5+t} \right)+\left( {n1} \right)\left( {2t} \right)=5+2ntt\). The equation then for this series’ partial sum is \({{S}_{n}}\,\,=\,\,\frac{n}{2}\left[ {\left( {5+t} \right)+\left( {5+2ntt} \right)} \right]=\frac{n}{2}\left( {2nt+10} \right)\). So \({{S}_{{20}}}\,\,=\,\frac{{20}}{2}\left( {2\left( {20} \right)t+10} \right)=400t+100\). 
Evaluate (find the sum):
\(33\frac{1}{2}4\,…\) to 50 terms  We can see that this is an arithmetic series, since the second term – the first = the third term – the second = –3.5 –(–3) = –.5; this is the d, or the common difference. The first term is –3, so \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d=3+\left( {n1} \right)\left( {.5} \right)=.5n2.5\). The equation then for this series’ partial sum is \({{S}_{n}}\,\,=\,\,\frac{n}{2}\left[ {3+\left( {.5n2.5} \right)} \right]=\frac{n}{2}\left( {.5n5.5} \right)\). So \({{S}_{{50}}}\,\,=\,\frac{{50}}{2}\left( {.5\left( {50} \right)5.5} \right)=762.5\) . 
Evaluate: \(\sum\limits_{{k=1}}^{n}{{\left( {8k+2} \right)}}\) Find the sum: \(\sum\limits_{{k=1}}^{{40}}{{\left( {8k+2} \right)}}\)  This series looks like 10, 18, 26, …. The first term is 10 and d = 8, so \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d=10+\left( {n1} \right)\left( 8 \right)=8n+2\). (Actually, we knew that from looking at the summation!) The equation then for this series’ partial sum is \({{S}_{n}}\,\,=\,\,\frac{n}{2}\left[ {10+\left( {8n+2} \right)} \right]=\frac{n}{2}\left( {8n+12} \right)=4{{n}^{2}}+6n\). So \({{S}_{{40}}}\,\,=\,\frac{{40}}{2}\left( {8\left( {40} \right)+12} \right)=6640\). 
Summation Notation
Before moving on to Geometric Series, we should get a little more practice with Summation Notation, since you’ll probably be seeing this in more advanced math classes. We will practice with writing out sums, and also expressing sums in summation notation:
Geometric Series
Just like we saw that an arithmetic series is the sum of an Arithmetic sequence, Geometric Series are the sum of Geometric Sequences.
For example, for the geometric sequence \({{a}_{n}}=2{{\left( 3 \right)}^{{n1}}}\), for n = 1 to 5, we have or 2, 6, 18, 54, 162. We would write this series as 2 + 6 + 18 + 54 + 162. Again, another way to write this is to use the summation formula \(\sum\limits_{{k=1}}^{5}{{\,2{{{\left( 3 \right)}}^{{k1}}}}}\). (The generic summation formula is \(\sum\limits_{{k=1}}^{n}{{\,{{a}_{1}}{{{\left( r \right)}}^{{k1}}}}}\)).
Here’s an example of why we might want to use a geometric series. Let’s say we know we are getting a 5% raise for the next five years, and we want to know the total amount of money we will make for these five years. We make $50,000 now.
Our common ratio will be 1.05 (do you see how we multiply this by what we make every year to get our new salary?). So we would need to get the sum of a geometric series of 5 terms (with common ratio 1.05) to see the total amount we’ll make in these five year, or .\(\sum\limits_{{k=1}}^{5}{{\,50000{{{\left( {1.05} \right)}}^{{k1}}}}}\)
Finite Geometric Series
For geometric series, we have different formulas, depending on whether the series ends at a certain point (finite), or goes on forever (infinite). It turns out that we may have an infinite geometric series that actually ends up at a number; this is said to converge to this number, since adding each term makes the series get closer to this number. We’ll talk about this below.
Here is the partial sum formula for the finite geometric series (one that doesn’t go to infinity):
This looks really complicated, but it’s really not too bad; if we are solving for the nth generic term \({{S}_{n}}\), we just plug in the first term (\({{a}_{1}}\)) and r (common ratio), and our answer has an “n” in it. If we are solving for an actual number, our answer won’t have an “n” in it. So it’s actually a little easier than the arithmetic series formula, since we don’t need to find \({{a}_{n}}\) first.
Note: we have to be careful to compute \({{r}^{n}}\) first, before subtracting this from 1, and then multiply this difference by \({{a}_{n}}\), and then divide the result by \(\left( {1r} \right)\). We can’t simply cancel out the \(\left( {1r} \right)\) and the \(\left( {1{{r}^{n}}} \right)\), unless n = 1.
Here are some problems:
Problem  Solution 
Find an equation for the nth partial sum \({{S}_{n}}\), and find \({{S}_{6}}\) in this geometric sequence: \({{a}_{1}}=2,\,\,\,\,r=4\)  We know the formula for the nth partial sum of a geometric series is \(\displaystyle {{S}_{n}}=\frac{{{{a}_{1}}\left( {1{{r}^{n}}} \right)}}{{1r}}\), so we have \(\displaystyle {{S}_{n}}=\frac{{2\left( {1{{4}^{n}}} \right)}}{{14}}=\frac{2}{3}\left( {1{{4}^{n}}} \right)\). (The series looks like 2 + 8 + 32 + …., which is \(\sum\limits_{{k=1}}^{n}{{2{{{\left( 4 \right)}}^{{k1}}}}}\)). To get the sum of the first 6 terms, we have \({{S}_{6}}=\,\frac{2}{3}\left( {1{{4}^{6}}} \right)=\frac{2}{3}\left( {4095} \right)=2730\). 
Find an equation for the nth partial sum \({{S}_{n}}\):
\(\displaystyle 81+\frac{1}{8}\frac{1}{{64}}…\)  We can see that this is a geometric series, since the second term/the first = the third term/the second = \(\frac{1}{8}\); this is r, or the common ratio.This series is \(\sum\limits_{{k=1}}^{n}{{8{{{\left( {\frac{1}{8}} \right)}}^{{k1}}}}}\). The equation for this series’ nth partial sum is \(\displaystyle {{S}_{n}}=\frac{{8\left( {1{{{\left( {\frac{1}{8}} \right)}}^{n}}} \right)}}{{1\left( {\frac{1}{8}} \right)}}=\frac{{8\left( {1{{{\left( {\frac{1}{8}} \right)}}^{n}}} \right)}}{{\frac{9}{8}}}=\frac{{64}}{9}\left( {1{{{\left( {\frac{1}{8}} \right)}}^{n}}} \right)\). Note that we have to be very careful with the parentheses; in the denominator, we can make the “– –“ into a “\(+\)”, but in the numerator, when the \(\frac{1}{8}\) is in the parentheses, we have to leave it. Try this geometric series for n = 2 (\({{S}_{n}}=7\)); it works! 
Find an equation for the nth partial sum \({{S}_{n}}\), and find \({{S}_{3}}\):
\(4{{\left( b \right)}^{{n1}}}\)  We know from the formula for a geometric sequence that \({{a}_{1}}=4\) and \(r=b\). Since the formula for the nth partial sum of a geometric series is \(\displaystyle {{S}_{n}}=\frac{{{{a}_{1}}\left( {1{{r}^{n}}} \right)}}{{1r}}\), we have \(\displaystyle {{S}_{n}}=\frac{{4\left( {1{{b}^{n}}} \right)}}{{1b}}\). The series looks like \(4+4b+4{{b}^{2}}+…\), or \(\sum\limits_{{k=1}}^{n}{{4{{{\left( b \right)}}^{{k1}}}}}\). To get the sum of the first 3 terms, we can factor the sum of two cubes to get:.\(\displaystyle \require{cancel} {{S}_{3}}\,\,=\,\,\frac{{4\left( {1{{b}^{3}}} \right)}}{{1b}}=\frac{{4\cancel{{\left( {1b} \right)}}\left( {1+b+{{b}^{2}}} \right)}}{{\cancel{{1b}}}}=4+4b+4{{b}^{2}}\). It works! 
Evaluate: \(\sum\limits_{{k=1}}^{5}{{8{{{\left( {\frac{1}{4}} \right)}}^{{k1}}}}}\)  We know from the summation formula for a geometric sequence that \({{a}_{1}}=8\) and \(r=\frac{1}{4}\). Since the formula for the nth partial sum of a geometric series is \(\displaystyle {{S}_{n}}=\frac{{{{a}_{1}}\left( {1{{r}^{n}}} \right)}}{{1r}}\), we have \(\displaystyle {{S}_{n}}=\frac{{8\left( {1{{{\left( {\frac{1}{4}} \right)}}^{n}}} \right)}}{{1\frac{1}{4}}}=\frac{{8\left( {1{{{\left( {\frac{1}{4}} \right)}}^{n}}} \right)}}{{\frac{3}{4}}}=\frac{{32}}{3}\left( {1{{{\left( {\frac{1}{4}} \right)}}^{n}}} \right)\). The series looks like \(8+2+\frac{1}{2}+\frac{1}{8}+\frac{1}{{32}}\). To get the sum of the first 5 terms, we have \({{S}_{5}}=\frac{{32}}{3}\left( {1{{{\left( {\frac{1}{4}} \right)}}^{5}}} \right)=10.65625=10\frac{{21}}{{32}}\). 
Infinite Geometric Series
A geometric series is infinite if it’s upper bound (the number on the top of the sigma sign) is infinity, or ∞. So an infinite series would look like this: \(\sum\limits_{{k=1}}^{\infty }{{2{{{\left( {.5} \right)}}^{{k1}}}}}\).
It turns out that the equation for an infinite series is much easier than that of a finite series, but there’s one caveat: we can only get an answer (meaning, the series converges), if the common ratio, or r , is between –1 and 1 (not including –1 and 1). This is the same thing as writing the restriction of r this way: \(\left r \right<1\). Note that the series diverges if \(\left r \right\ge 1\); we can’t find an answer to the infinite series.
This makes sense since if r is a fraction, we’ll be adding a term that is getting smaller and smaller, so we should finally arrive at a number if we theoretically go to infinity (called the “limit”).
Here is the formula:
Here are some problems; determine whether the geometric series converges or diverges; if it converges, find the sum:
Summary of Formulas for Sequences and Series
Here is a summary of the main formulas for Sequences and Series:
\(\begin{array}{l}{{a}_{1}}=\,\,\text{first term}\\{{a}_{n}}=\,\,n\text{th term}\end{array}\)  Arithmetic d = common difference  Geometric r = common ratio 
Sequences 
\({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d\)
(Recursive: \({{a}_{n}}=a,\,\,\,\,\,{{a}_{n}}={{a}_{{n1}}}+d\))
 \({{a}_{n}}={{a}_{1}}{{\left( r \right)}^{{n1}}}\)
(Recursive: \({{a}_{n}}=a,\,\,\,\,\,{{a}_{n}}=r{{a}_{{n1}}}\) ) 
Finite Series (Sum) 
\(\displaystyle {{S}_{n}}=\frac{n}{2}\left( {{{a}_{1}}+{{a}_{n}}} \right)\,\,\,=\,\,\,\frac{n}{2}\left( {2{{a}_{1}}+\left( {n1} \right)d} \right)\)
 \(\displaystyle {{S}_{n}}=\frac{{{{a}_{1}}\left( {1{{r}^{n}}} \right)}}{{1r}}\) 
Infinite Series (Sum)  Not Applicable 
\(\displaystyle {{S}_{\infty }}=\frac{{{{a}_{1}}}}{{1r}},\,\,\,\,\,\,\,\left r \right<1\)

Applications of Sequences and Series
Sequence and Series are very useful in many applications; in fact, with geometric sequences, especially when we’re dealing with growth or decay (like with money), we’ll see that they equations look a lot like some of the exponential equations we worked with here in the Exponential Functions section.
The main thing to remember about word problems with sequences and series is that when we want an amount for a single thing, such as a particular row, year, for example, we use a sequence. When we want to know a total amount, such as money or rows, we want to use a series (which is a sum).
Here are some problems:
Problem:
A stadium with 75 rows has 10 seats in the first row, 15 in the second, and so on. How many seats are in the top row? How many total seats are in the stadium?
Solution:
We can see that an arithmetic sequence models this situation (since we are adding 5 seats to each successive row), and when we are adding up all the seats in the stadium, we are dealing with an arithmetic series.
To get how many seats are in the top row, we have \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d\), where \({{a}_{1}}=10\), and d = 5. We have \({{a}_{{75}}}=10+\left( {751} \right)5=380\), so there are 380 seats in the top row.
To get the total number of seats, we’ll use \({{S}_{n}}=\frac{n}{2}\left( {{{a}_{1}}+{{a}_{n}}} \right)\), or \({{S}_{{75}}}=\frac{{75}}{2}\left( {10+380} \right)=14625\), so there are a total of 14625 seats. See how useful these equations could be, if you had to put in an order in for these seats?
Problem:
For the first year of a new job, you make $60,000. You get an annual raise each year of 3%. Find out how much you’ll make in your fifth year, and also the total amount of money you will make in 5 years.
Solution
Let’s first put some real numbers in to see what type of sequence or series this is. If you make $60,000 the first year, and you get a 3% raise for the second year, you will make 60000 + (.03)(60000), or $61,800. (This is the same as multiplying 60000 by 1.03). To get how much you’ll make in the third year, multiply $61,800 by 1.03 again. So we have a geometric sequence with r = 1.03.
To find out how much money you’ll make in the fifth year, we’ll use the equation \({{a}_{n}}={{a}_{1}}{{\left( r \right)}^{{n1}}}\), with n = 5, \({{a}_{1}}\) = 600000, and r = 1.03. We’ll have \({{a}_{5}}=60000{{\left( {1.03} \right)}^{{51}}}=67530.53\). So you will make $67,530.53 in year 5. This makes sense, since we know it needs to be more than $60,000.
To find out how much you will make in the first 5 years, we need the partial sum formula for a geometric series, which is \({{S}_{n}}\,\,=\,\,\frac{{{{a}_{1}}\left( {1{{r}^{n}}} \right)}}{{1r}}\). We have \({{S}_{5}}\,\,=\,\,\frac{{60000\left( {1{{{\left( {1.03} \right)}}^{5}}} \right)}}{{11.03}}=318548.15\). So you’ll make $318,548.15 in the first 5 years!
Problem:
A pendulum swings through an arc (distance of the curve) of 5 feet initially. For the second swing, the length of the arc is .85 the length of the previous swing, and so on. On which swing will the length of the arc be less than 2 feet?
Solution:
We can see that we have a geometric sequence here, since we are multiplying the length of the arc by .85 with each swing. Since the initial swing is 5 feet, we have \({{a}_{n}}={{a}_{1}}{{\left( r \right)}^{{n1}}}\), or \({{a}_{n}}=5{{\left( {.85} \right)}^{{n1}}}\).
Since the question asks when (which swing) the length of the arc will be less than 2 feet, we need to use the equation \(5{{\left( {.85} \right)}^{{n1}}}=2\) to see what n is when the swing of the arc is 2 feet. The best way to solve this is to use logs (you could also use “guess and check”); we have \(5{{\left( {.85} \right)}^{{n1}}}=2;\,\,\,{{\left( {.85} \right)}^{{n1}}}=\frac{2}{5};\,\,\,\left( {n1} \right)\ln \left( {.85} \right)=\ln \left( {\frac{2}{5}} \right);\,\,n\approx 6.6\).
So when the length of the arc is 2, it will be after the 6^{th} swing, since n has to be an integer. So the swing will have an arc length of less than 2 on the 7^{th} swing. Tricky!
Problem:
How much (what fraction) of the square will eventually be shaded if the shading pattern (in the shapes of squares) is continued indefinitely from the top left to the bottom right?
Solution:
It looks like the first shaded square is one quarter of the whole square, and the second shaded square is one quarter of that, and so on. Since the shading theoretically goes on indefinitely, this will be a infinite geometric series, with \({{a}_{1}}=\frac{1}{4}\), and \(r=\frac{1}{4}\) (see how second/first = \(\frac{1}{4}\) )? So we have \({{S}_{\infty }}=\frac{{{{a}_{1}}}}{{1r}}=\frac{{\left( {\frac{1}{4}} \right)}}{{1\left( {\frac{1}{4}} \right)}}=\frac{{\left( {\frac{1}{4}} \right)}}{{\left( {\frac{3}{4}} \right)}}=\frac{1}{3}\).
Problem:
The 5^{th} term of an arithmetic sequence is –2 and the sum of the first 10 terms of the sequence is –35. Find the sequence.
Solution:
Let’s write down what we know, using the formulas we know. Since the 5^{th} term of the arithmetic sequence is –2, we know \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d\), or \(2={{a}_{1}}+\left( {51} \right)d\), so \(2={{a}_{1}}+4d\). We have two unknown variables, so we’ll need another equation to solve.
Since the sum of the first 10 terms is –35, we have \({{S}_{n}}=\frac{n}{2}\left( {{{a}_{1}}+{{a}_{n}}} \right)\), or \(35=\frac{{10}}{2}\left( {{{a}_{1}}+{{a}_{{10}}}} \right)\), or \(35=\frac{{10}}{2}\left[ {{{a}_{1}}+{{a}_{1}}+\left( {101} \right)d} \right]\), or \(35=10{{a}_{1}}+45d\).
So now we have two variables and two unknowns, so we have a solvable system. We can solve with substitution:
\(\begin{array}{c}2={{a}_{1}}+4d;\,\,\,\,\,\,\,\,\,{{a}_{1}}=24d\\35=10{{a}_{1}}+45d;\,\,\,\,\,\,35=10\left( {24d} \right)+45d;\,\,\,\,\,\,35=2040d+45d;\,\,\,\,\,\,\,15=5d;\,\,\,\,\,\,d=3\\{{a}_{1}}=24\left( {3} \right);\,\,\,\,\,{{a}_{1}}=\,\,10\end{array}\)
So this sequence is \({{a}_{n}}=10+\left( {n1} \right)\left( {3} \right)=3n+13\). Try it – it works!
Sequences and Sums on the Graphing Calculator
We can use the TI Graphing Calculator to create sequences and determine the sum of sequences (series).
We can use SEQ and SUM in the Catalog list, or in the 2^{nd} STAT (LIST) OPS 5 or seq and 2^{nd} STAT MATH 5 or sum, as in the following. Note that the calculator will either have you fill in the steps or (if you have an older operating system), you have to enter the arguments on one line.
You first enter the expression, using the X,T,θ,n button, the variable (hit X,T,θ,n again), starting and ending values of the variable, and (optional) 1 if you want all the values – which you probably will.
With the newer operating systems, when you use sum, you can get a summation symbol where you can enter the variable, the beginning and ending values, and the expression, like in the last screen. (You can also use MATH summation (MATH 0) or the shortcut ALPHA WINDOW 2 to do this.)
Note that if you want the sum of an infinite series, you can put a large number in the upper bound, such as 999, and you should arrive at the solution, or pretty close!
You can even put a sequence in a LIST, like \({{L}_{1}}\), by using the seq command and then using STO and then the name of the list:
Learn these rules, and practice, practice, practice!
For Practice: Use the Mathway widget below to try a Sequences and Series problem. Click on Submit (the blue arrow to the right of the problem) and click on Identify the Sequence to see the answer.
You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.
If you click on Tap to view steps, or Click Here, you can register at Mathway for a free trial, and then upgrade to a paid subscription at any time (to get any type of math problem solved!).
On to Binomial Expansion – you are ready!