Definition 5.1.1.
A sequence is an ordered list.
Section 5.1 Sequences
In this section we look at the mathematical concept of sequences. Although many sequences themselves are straightfoward, such as \(2, 4, 6, 8, \ldots\text{,}\) we need to introduce notation and terminology for working with general sequences.
We use the notation \(a_1, a_2, a_3, \ldots, a_k, \ldots\) for a general sequence.
Each \(a_k\) is called a term in the sequence. The subscript \(k\) is called the index. The index will be an integer, and almost always a nonnegative integer. The first term \(a_1\) (or sometimes \(a_0\)) is called the initial term. The term \(a_k\) is called the \(k^{th}\) term. It is also often called the general term of the sequence.
Example 5.1.2. Sequences.
Consider the sequence \(2, 4, 6, 8, 10, \ldots\text{.}\) The initial term is \(a_1=2\text{.}\) The \(k^{th}\) term is \(a_k=2k\text{.}\)
We need to be careful with subscripts. For example, \(a_{4+1}=a_{5}=10\text{,}\) but \(a_{4}+1=8+1=9\text{.}\) If we add 1 to the index, we get the next term, which is not the same as adding 1 to the term.
Example 5.1.3. Explicitly Defined Sequences.
We can define a sequence by giving the general term.
Let \(a_k=2^k, k\geq 0\text{.}\) Give the first five terms of the sequence.
Answer 1.
Let \(a_k=2^k, k\geq 0\text{.}\) Give the \(k+1\) term of the sequence.
Answer 2.
\(a_{k+1}=2^{k+1}\)
Let \(a_k=\frac{1}{k+1}, k\geq 1\text{.}\) Give the first five terms of the sequence.
Answer 3.
1/2, 1/3, 1/4, 1/5, 1/6
Let \(a_k=\frac{1}{k+1}, k\geq 1\text{.}\) Give the \(k+1\) term of the sequence.
Answer 4.
\(a_{k+1}=\frac{1}{k+2}\)
Activity 5.1.1.
Consider the sequence \(a_k=(-1)^k\) for \(k\geq 0\text{.}\)
(a)
Write the first 5 terms of the sequence.
(b)
What is the initial term?
Activity 5.1.2.
Consider the sequence \(a_k=\frac{1}{k-1}\) for \(k\geq 3\text{.}\)
(a)
Write the first 5 terms of the sequence.
(b)
What is the initial term?
Activity 5.1.3.
Consider the sequence \(0, 1, -2, 3, -4, 5, \ldots\text{.}\) Find a general formula for the \(k\)th term, \(a_k\text{.}\)
We are going to look at many examples where we want to add terms in a sequence. The following notation will be helpful when working with sums.
Summation Notation.
We can write a sum using sigma or summation notation:
\begin{equation*}
a_1+a_2+\cdots +a_n=\sum_{k=1}^{n}a_k.
\end{equation*}
We read \(\sum_{k=1}^{n}a_k\) as “the sum of \(a_k\) from \(k=1\) to \(n\text{.}\)”
Example 5.1.4. Summation Notation.
Find \(\sum_{k=1}^4 k\text{.}\)
Answer 1.
\(1+2+3+4=10\)
Find \(\sum_{k=1}^5 k^2\text{.}\)
Answer 2.
\(1^2+2^2+3^2+4^2+5^2=55\)
Find \(\sum_{k=1}^n k^2\text{.}\)
Answer 3.
\(1^2+2^2+3^2+\cdots+n^2\)
Find \(\sum_{k=2}^2 k^2\text{.}\)
Answer 4.
\(2^2=4\)
Note, we can write the sum of only the \(m^{th}\) term, \(\sum_{k=m}^{m}a_k=a_m\text{.}\)
Activity 5.1.4.
Consider the sum \(\sum_{k=1}^{5}(2k-1)\text{.}\) Write out the summation and find the sum.Activity 5.1.5.
Consider the sum \(\sum_{k=1}^{n}\frac{1}{k}\text{.}\)
(a)
Write out the summation.
(b)
Write out the summation for \(\sum_{k=1}^{n+1}\frac{1}{k}\text{.}\) How do (a) and (b) differ?
(c)
Write out the summation for \(\sum_{k=0}^{n}\frac{1}{k+1}\text{.}\) Is this the same as either of the previous sums?
Activity 5.1.6.
Consider the sum \(\sum_{k=1}^{n}\frac{1}{k(k+1)}\text{.}\)
(a)
Write out the summation.
(b)
Write out the summation for \(\sum_{k=1}^{n+1}\frac{1}{k(k+1)}\text{.}\) How do (a) and (b) differ?
Just as we can add several terms of a sequence, the following notation alllows us to multiply several terms of a sequence using product notation:
\begin{equation*}
a_1\cdot a_2\cdot a_3\cdots a_n=\prod_{k=1}^{n}a_k.
\end{equation*}
Example 5.1.5. Product Notation.
Find \(\prod_{k=1}^4 k\text{.}\)
Answer 1.
\(1\cdot2\cdot3\cdot 4=24\)
Find \(\prod_{k=1}^3 k^2\text{.}\)
Answer 2.
\(1^2\cdot2^2\cdot3^2=36\)
Activity 5.1.7.
Write out the following products.
(a)
\(\prod_{k=1}^{5}(2k-1)\)
(b)
\(\prod_{k=1}^{n+1}(2k)\)
Recall, we defined \(n\) factorial in Definition 4.3.7: \(n!=(n)(n-1)\cdots(2)(1)\text{.}\) We also need to define \(0!=1\text{.}\)
The following properties are helpful when working with sums and products.
Properties of Sums and Products.
- \(\displaystyle \sum_{k=m}^{n}a_k+\sum_{k=m}^{n}b_k=\sum_{k=m}^{n}(a_k+b_k)\)
- \(\displaystyle c\sum_{k=m}^{n}a_k=\sum_{k=m}^{n}(ca_k)\)
- \(\displaystyle \biggl(\prod_{k=m}^{n}a_k\biggr)\biggl(\prod_{k=m}^{n}b_k\biggr)=\prod_{k=m}^{n}(a_k\cdot b_k)\)
Activity 5.1.8.
Prove \(\sum_{k=1}^{n}a_k + \sum_{k=1}^{n}b_k=\sum_{k=1}^{n}(a_k+b_k)\text{.}\)
Hint.
Try writing out the sum rather than using summation notation.
Definition 5.1.6.
The number of subsets of size \(r\) that can be chosen from a set of \(n\) elements is \(n\) choose \(r\text{.}\) Notation \(\binom{n}{r}\), read “\(n\) choose \(r\text{.}\)”
We can calculate the number of sets of \(r\) objects chosen from \(n\) objects with the following formula:
\begin{equation*}
\binom{n}{r}=\frac{n!}{r!(n-r)!}.
\end{equation*}
Example 5.1.7. Finding \(n\) choose \(r\).
Calculate \(\binom{5}{3}\text{.}\)
Answer 1.
\(\frac{5!}{3!2!}=10\)
Calculate \(\binom{5}{1}\text{.}\)
Answer 2.
\(\frac{5!}{1!4!}=5\)
Calculate \(\binom{5}{4}\text{.}\)
Answer 3.
\(\frac{5!}{4!1!}=5\)
Calculate \(\binom{5}{0}\text{.}\)
Answer 4.
\(\frac{5!}{0!5!}=1\)
Activity 5.1.9.
Find \({6 \choose 3}\) and \({6 \choose 0}\text{.}\)
When we get to mathematical induction in the next section, it will be important that we can work with summations when we want to add “the \(n+1\) term” to a summation. In particular, the following observation is useful:
\begin{equation*}
\biggl(\sum_{k=1}^{n}a_{k}\biggr)+a_{n+1}=\sum_{k=1}^{n+1}a_k.
\end{equation*}
We should also note that there are often multiple ways to write the same sum.
Example 5.1.8. Writing a Sum in Two Different Ways.
Consider the sum \(1^2+2^2+3^2\text{.}\) Depending on how we index the sum, we can write it in different ways.
If we index from \(k=1\) to 3, we have \(\sum_{k=1}^3k^2=1^2+2^2+3^2\text{.}\)
If we index from \(k=2\) to 4, we have \(\sum_{k=2}^4(k-1)^2=1^2+2^2+3^2\text{.}\)
Reading Questions Check Your Understanding
1.
Write the terms of \(\frac{k+1}{k+3}, 0\leq k\leq 4\text{.}\)
2.
Write the terms of \(\frac{(-1)^k}{2k}, 1\leq k\leq 6\text{.}\)
3.
Write the terms of \(\frac{(-1)^kk}{2}, 1\leq k\leq 4\text{.}\)
4.
Write out the summation notation as a sum of terms: \(\sum_{k=1}^4 2k-1\)
5.
Write out the summation notation as a sum of terms: \(\sum_{k=1}^n k^3\)
6.
Write out the summation notation as a sum of terms: \(\sum_{k=1}^{n+1} k^3\)
7.
Write out the summation notation as a sum of terms: \(\sum_{k=1}^4 \frac{(-1)^k}{k}\)
8.
Write out the summation notation as a sum of terms: \(\sum_{k=1}^n \frac{(-1)^k}{k}\)
9.
Write out the summation notation as a sum of terms: \(\sum_{k=1}^{n+1} \frac{(-1)^k}{k}\)
Exercises Exercises
1.
Find an explicit formula for the following sequences with the given initial terms.
- \(\displaystyle \frac{1}{3}, \frac{4}{9},\frac{9}{27}, \frac{16}{81},\frac{25}{243}, \frac{36}{729}\)
- \(\displaystyle 3, 6, 12, 24, 48, 96\)
2.
Compute the given product or sum.
- \(\displaystyle \prod_{k=2}^{4}{k^2}\)
- \(\displaystyle \prod_{k=2}^{2}{\Bigl(1-\frac{1}{k}\Bigr)}\)
- \(\displaystyle \sum_{k=-1}^{1}{(k^2+3)}\)
3.
Write out the sum in expanded form.
- \(\displaystyle \sum_{j=1}^{n}{j(j+1)}\)
- \(\displaystyle \sum_{i=1}^{k+1}{i(i!)}\)
4.
Rewrite by separating off the final term: \(\sum_{i=1}^{k+1}{i(i!)}\)
5.
Write using product notation:
\begin{equation*}
(2^2-1)\cdot(3^2-1)\cdot(4^2-1).
\end{equation*}
6.
Write using summation notation:
\begin{equation*}
1^3+2^3+3^3+\cdots+n^3.
\end{equation*}
7.
Transform the sum by making the change of variable \(j=i-1\text{:}\)
\begin{equation*}
\sum_{i=1}^{n+1}\frac{(i-1)^2}{i\cdot n}.
\end{equation*}
8.
Simplify.
- \(\displaystyle \frac{((n+1)!)^2}{(n!)^2}\)
- \(\displaystyle \frac{n!}{(n-k+1)!}\)
9.
Compute.
- \(\displaystyle {3 \choose 0}\)
- \(\displaystyle {n \choose n-1}\)
- \(\displaystyle {n+1 \choose n-1}\)
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