Writing the sum of long lists of numbers that have a specific pattern is not very efficient. Summation notation allows you to use the pattern and the number of terms to represent the same sum in a much more concise way. How can you use sigma notation to represent the following sum?

\begin{align*}1+4+9+16+25+\cdots+144\end{align*}

#### Watch This

http://www.youtube.com/watch?v=0L0rU17hHuM James Sousa: Find a Sum Written in Summation/Sigma Notation

#### Guidance

A
**
series
**
is a sum of a sequence. The Greek capital letter sigma is used for summation notation because it stands for the letter
@$\begin{align*}S\end{align*}@$
as in sum.

Consider the following general sequence and note that the subscript for each term is an index telling you the term number.

@$\begin{align*}a_1, a_2, a_3, a_4, a_5\end{align*}@$

When you write the sum of this sequence in a series, it can be represented as a sum of each individual term or abbreviated using a capital sigma.

@$\begin{align*}a_1+a_2+a_3+a_4+a_5=\sum\limits_{i=1}^5 a_i\end{align*}@$

The three parts of sigma notation that you need to be able to read are the argument, the lower index and the upper index. The argument, @$\begin{align*}a_i\end{align*}@$ , tells you what terms are added together. The lower index, @$\begin{align*}i=1\end{align*}@$ , tells you where to start and the upper index, 5, tells you where to end. You should practice reading and understanding sigma notation because it is used heavily in Calculus.

**
Example A
**

Write out all the terms of the series.

@$\begin{align*}\sum\limits_{k=4}^8 2k\end{align*}@$

**
Solution:
**

@$\begin{align*}\sum\limits_{k=4}^8 2k=2\cdot4+2\cdot5+2\cdot6+2\cdot7+2\cdot8\end{align*}@$

**
Example B
**

Write the sum in sigma notation: @$\begin{align*}2+3+4+5+6+7+8+9+10\end{align*}@$

**
Solution:
**

@$\begin{align*}2+3+4+5+6+7+8+9+10=\sum\limits_{i=2}^{10} i\end{align*}@$

**
Example C
**

Write the sum in sigma notation.

@$\begin{align*}\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}\end{align*}@$

**
Solution:
**

@$\begin{align*}\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}=\sum\limits_{i=1}^7 \frac{1}{i^2}\end{align*}@$

**
Concept Problem Revisited
**

The hardest part when first using sigma representation is determining how each pattern generalizes to the @$\begin{align*}k^{th}\end{align*}@$ term. Once you know the @$\begin{align*}k^{th}\end{align*}@$ term, you know the argument of the sigma. For the sequence creating the series below, @$\begin{align*}a_k=k^2\end{align*}@$ . Therefore, the argument of the sigma is @$\begin{align*}i^2\end{align*}@$ .

@$\begin{align*}1+4+9+16+25+\cdots+144=1^2+2^2+3^2+4^2+\cdots 12^2=\sum\limits_{i=1}^{12} i^2\end{align*}@$

#### Vocabulary

**
Sigma notation
**
is also known as

**and is a way to represent a sum of numbers. It is especially useful when the numbers have a specific pattern or would take too long to write out without abbreviation.**

*summation notation*#### Guided Practice

1. Write out all the terms of the sigma notation and then calculate the sum.

@$\begin{align*}\sum\limits_{k=0}^4 3k-1\end{align*}@$

2. Represent the following infinite series in summation notation.

@$\begin{align*}\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots\end{align*}@$

3. Is there a way to represent an infinite product? How would you represent the following product?

@$\begin{align*}1\cdot\sin\left(\frac{360}{3}\right)\cdot\sin\left(\frac{360}{4}\right)\cdot\sin\left(\frac{360}{5}\right)\cdot\sin\left(\frac{360}{6}\right)\cdot\sin\left(\frac{360}{7}\right)\cdot\ldots\end{align*}@$

**
Answers:
**

1. @$$\begin{align*}\sum\limits_{k=0}^4 3k-1&=(3\cdot0-1)+(3\cdot1-1)+(3\cdot2-1)+(3\cdot3-1)+(3\cdot4-1)\\ &=-1+2+5+8+11\end{align*}@$$

2. There are an infinite number of terms in the series so using an infinity symbol in the upper limit of the sigma is appropriate.

@$\begin{align*}\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\cdots=\sum\limits_{i=1}^\infty \frac{1}{2^i}\end{align*}@$

3. Just like summation uses a capital Greek letter for @$\begin{align*}S\end{align*}@$ , product uses a capital Greek letter for @$\begin{align*}P\end{align*}@$ which is the capital form of @$\begin{align*}\pi\end{align*}@$ .

@$\begin{align*}1\cdot\sin\left(\frac{360}{2\cdot3}\right)\cdot\sin\left(\frac{360}{2\cdot4}\right)\cdot\sin\left(\frac{360}{2\cdot5}\right)\cdot\sin\left(\frac{360}{2\cdot6}\right)\cdot\sin\left(\frac{360}{2\cdot7}\right)\cdot\ldots=\prod \limits_{i=3}^{\infty} \sin \left(\frac{360}{2 \cdot i} \right)\end{align*}@$

This infinite product is the result of starting with a circle of radius 1 and inscribing a regular triangle inside the circle. Then you inscribe a circle inside the triangle and a square inside the new circle. The shapes alternate being inscribed within each other as they are nested inwards: circle, triangle, circle, square, circle, pentagon, ... The question that this calculation starts to answer is whether this process reduces to a number or to zero.

#### Practice

For 1-5, write out all the terms of the sigma notation and then calculate the sum.

1. @$\begin{align*}\sum\limits_{k=1}^5 2k-3\end{align*}@$

2. @$\begin{align*} \sum\limits_{k=0}^8 2^k\end{align*}@$

3. @$\begin{align*}\sum\limits_{i=1}^4 2 \cdot 3^i\end{align*}@$

4. @$\begin{align*}\sum\limits_{i=1}^{10} 4i-1\end{align*}@$

5. @$\begin{align*}\sum\limits_{i=0}^4 2\cdot\left(\frac{1}{3}\right)^i\end{align*}@$

Represent the following series in summation notation with a lower index of 0.

6. @$\begin{align*}1+4+7+10+13+16+19+22\end{align*}@$

7. @$\begin{align*} 3+5+7+9+11\end{align*}@$

8. @$\begin{align*}8+7+6+5+4+3+2+1\end{align*}@$

9. @$\begin{align*}5+6+7+8\end{align*}@$

10. @$\begin{align*}3+6+12+24+48+\cdots\end{align*}@$

11. @$\begin{align*}10+5+\frac{5}{2}+\frac{5}{4}\end{align*}@$

12. @$\begin{align*}4-8+16-32+64\cdots\end{align*}@$

13. @$\begin{align*}2+4+6+8+\cdots\end{align*}@$

14. @$\begin{align*}\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\cdots\end{align*}@$

15. @$\begin{align*}\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\frac{2}{81}+\cdots\end{align*}@$