Section 4.1 Summation Notation
In this section we look at summation notation, which is used to represent general sums, even infinite sums. Before we add terms together, we need some notation for the terms themselves.
A sequence is an ordered list, \(a_1, a_2, a_3, \ldots, a_k, \ldots\text{.}\)
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.
Example 4.1.2. 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 1.1, 2, 4, 8, 16
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 4.1.1. An Alternating Sequence.
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 4.1.2. A Fractional Sequence.
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 4.1.3. Another Alternating Sequence.
Consider the sequence \(0, 1, -2, 3, -4, 5, \ldots\text{.}\) Find a general formula for the \(k\)th term, \(a_k\text{.}\)
Now that we have general forms for sequences, we look at many examples where we add terms in a sequence.
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 4.1.3. 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, as we can see in Example 4.1.3, \(\sum_{k=m}^{m}a_k=a_m\text{.}\)
Activity 4.1.4. Using Summation Notation.
Consider the sum \(\sum_{k=1}^{5}(2k-1)\text{.}\) Write out the summation and find the sum.Activity 4.1.5. A Shifted Sum.
Consider the sum \(\sum_{k=1}^{n}\frac{1}{k}\text{.}\)
(a)
Write out the summation.
(b)
Now write out the summation for \(\sum_{k=1}^{n+1}\frac{1}{k}\text{.}\) How do (a) and (b) differ?
(c)
Now 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 4.1.6. Adding Another Term.
Consider the sum \(\sum_{k=1}^{n}\frac{1}{k(k+1)}\text{.}\)
(a)
Write out the summation.
(b)
Now 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, product notation allows us to multiply several terms of a sequence.
\begin{equation*}
a_1\cdot a_2\cdot a_3\cdots a_n=\prod_{k=1}^{n}a_k\text{.}
\end{equation*}
Example 4.1.4. 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 4.1.7. Products.
Write out the following products.
(a)
\(\prod_{k=1}^{5}(2k-1)\)
(b)
\(\prod_{k=1}^{n+1}(2k)\)
A particularly useful product is \(n\) factorial,
\begin{equation*}
n!=(n)(n-1)\cdots(2)(1).
\end{equation*}
We also need to define \(0!=1\text{.}\)
Properties of Sums and Products.
The following are useful properties when working with summation and product notation.
- \(\displaystyle \sum_{k=m}^{n}a_k+\sum_{k=m}^{n}b_k=\sum_{k=m}^{n}(a_k+b_k)\)
- \(\displaystyle c\sum_{k=m}^{n}a_k=\sum_{k=m}^{n}(ca_k)\)
- \(\displaystyle \biggl(\prod_{k=m}^{n}a_k\biggr)\biggl(\prod_{k=m}^{n}b_k\biggr)=\prod_{k=m}^{n}(a_k\cdot b_k)\)
Activity 4.1.8. Summing Two Sums.
Prove \(\sum_{k=1}^{n}a_k + \sum_{k=1}^{n}b_k=\sum_{1}^{n}(a_k+b_k)\text{.}\)
Hint.
Try writing out the sum rather than using summation notation.
When we get to mathematical induction in Section 4.2, it will be important that we can work with summations where 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 4.1.5. 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 \(\sum_{k=2}^4(k-1)^2=1^2+2^2+3^2\text{.}\)
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.
Observe that
\begin{align*}
\frac{1}{1\cdot 3}\amp =\frac{1}{3}\\
\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}\amp =\frac{2}{5}\\
\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}\amp =\frac{3}{7}\\
\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}\amp =\frac{4}{9}
\end{align*}
Find a general formula for \(\sum_{i=1}^{n}\frac{1}{(2i-1)(2i+1)}\text{.}\)
10.
Evaluate the sum \(\sum_{i=1}^{n}\frac{i}{(i+1)!}\) for all \(n=1, 2, 3, 4, 5\text{.}\) Find a formula for the sum for a general \(n\text{.}\)
