### Definition 5.1.1.

A sequence is an ordered list.

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.

A sequence is an ordered list.

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.

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.

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.

1, 2, 4, 8, 16

Let \(a_k=2^k, k\geq 0\text{.}\) Give the \(k+1\) term of the sequence.

\(a_{k+1}=2^{k+1}\)

Let \(a_k=\frac{1}{k+1}, k\geq 1\text{.}\) Give the first five terms of the sequence.

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.

\(a_{k+1}=\frac{1}{k+2}\)

Consider the sequence \(a_k=(-1)^k\) for \(k\geq 0\text{.}\)

Write the first 5 terms of the sequence.

What is the initial term?

Consider the sequence \(a_k=\frac{1}{k-1}\) for \(k\geq 3\text{.}\)

Write the first 5 terms of the sequence.

What is the initial term?

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.

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 \(n\text{.}\)”

Find \(\sum_{k=1}^4 k\text{.}\)

\(1+2+3+4=10\)

Find \(\sum_{k=1}^5 k^2\text{.}\)

\(1^2+2^2+3^2+4^2+5^2=55\)

Find \(\sum_{k=1}^n k^2\text{.}\)

\(1^2+2^2+3^2+\cdots+n^2\)

Find \(\sum_{k=2}^2 k^2\text{.}\)

\(2^2=4\)

Note, we can write the sum of only the \(m^{th}\) term, \(\sum_{k=m}^{m}a_k=a_m\text{.}\)

Consider the sum \(\sum_{k=1}^{n}\frac{1}{k}\text{.}\)

Write out the summation.

Write out the summation for \(\sum_{k=1}^{n+1}\frac{1}{k}\text{.}\) How do (a) and (b) differ?

Write out the summation for \(\sum_{k=0}^{n}\frac{1}{k+1}\text{.}\) Is this the same as either of the previous sums?

Consider the sum \(\sum_{k=1}^{n}\frac{1}{k(k+1)}\text{.}\)

Write out the summation.

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*}

Find \(\prod_{k=1}^4 k\text{.}\)

\(1\cdot2\cdot3\cdot 4=24\)

Find \(\prod_{k=1}^3 k^2\text{.}\)

\(1^2\cdot2^2\cdot3^2=36\)

Write out the following products.

\(\prod_{k=1}^{5}(2k-1)\)

\(\prod_{k=1}^{n+1}(2k)\)

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.

- \(\displaystyle \sum_{k=m}^{n}a_k+\sum_{k=m}^{n}b_k=\sum_{k=m}^{n}(a_k+b_k)\)
- \(\displaystyle c\sum_{k=m}^{n}a_k=\sum_{k=m}^{n}(ca_k)\)
- \(\displaystyle \biggl(\prod_{k=m}^{n}a_k\biggr)\biggl(\prod_{k=m}^{n}b_k\biggr)=\prod_{k=m}^{n}(a_k\cdot b_k)\)

Prove \(\sum_{k=1}^{n}a_k + \sum_{k=1}^{n}b_k=\sum_{k=1}^{n}(a_k+b_k)\text{.}\)

Try writing out the sum rather than using summation notation.

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*}

Calculate \(\binom{5}{3}\text{.}\)

\(\frac{5!}{3!2!}=10\)

Calculate \(\binom{5}{1}\text{.}\)

\(\frac{5!}{1!4!}=5\)

Calculate \(\binom{5}{4}\text{.}\)

\(\frac{5!}{4!1!}=5\)

Calculate \(\binom{5}{0}\text{.}\)

\(\frac{5!}{0!5!}=1\)

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.

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 \(\sum_{k=2}^4(k-1)^2=1^2+2^2+3^2\text{.}\)

Write the terms of \(\frac{k+1}{k+3}, 0\leq k\leq 4\text{.}\)

Write the terms of \(\frac{(-1)^k}{2k}, 1\leq k\leq 6\text{.}\)

Write the terms of \(\frac{(-1)^kk}{2}, 1\leq k\leq 4\text{.}\)

Write out the summation notation as a sum of terms: \(\sum_{k=1}^4 2k-1\)

Write out the summation notation as a sum of terms: \(\sum_{k=1}^n k^3\)

Write out the summation notation as a sum of terms: \(\sum_{k=1}^{n+1} k^3\)

Write out the summation notation as a sum of terms: \(\sum_{k=1}^4 \frac{(-1)^k}{k}\)

Write out the summation notation as a sum of terms: \(\sum_{k=1}^n \frac{(-1)^k}{k}\)

Write out the summation notation as a sum of terms: \(\sum_{k=1}^{n+1} \frac{(-1)^k}{k}\)

Find an explicit formula for the following sequences with the given initial terms.

- \(\displaystyle \frac{1}{3}, \frac{4}{9},\frac{9}{27}, \frac{16}{81},\frac{25}{243}, \frac{36}{729}\)
- \(\displaystyle 3, 6, 12, 24, 48, 96\)

Compute the given product or sum.

- \(\displaystyle \prod_{k=2}^{4}{k^2}\)
- \(\displaystyle \prod_{k=2}^{2}{\Bigl(1-\frac{1}{k}\Bigr)}\)
- \(\displaystyle \sum_{k=-1}^{1}{(k^2+3)}\)

Write out the sum in expanded form.

- \(\displaystyle \sum_{j=1}^{n}{j(j+1)}\)
- \(\displaystyle \sum_{i=1}^{k+1}{i(i!)}\)

Rewrite by separating off the final term: \(\sum_{i=1}^{k+1}{i(i!)}\)

Write using product notation:

\begin{equation*}
(2^2-1)\cdot(3^2-1)\cdot(4^2-1).
\end{equation*}

Write using summation notation:

\begin{equation*}
1^3+2^3+3^3+\cdots+n^3.
\end{equation*}

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*}

Simplify.

- \(\displaystyle \frac{((n+1)!)^2}{(n!)^2}\)
- \(\displaystyle \frac{n!}{(n-k+1)!}\)

Compute.

- \(\displaystyle {3 \choose 0}\)
- \(\displaystyle {n \choose n-1}\)
- \(\displaystyle {n+1 \choose n-1}\)

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