3.6 Summation Notation

Summation Notation

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Find the Partial Sum

Sometimes in applications, rather than just list the terms, it is important for us to add the terms of a sequence. Rather than just connect the terms with plus signs, we can use summation notation.

For example, $a_{1} + a_{2} + a_{3} + a_{4} + a_{5}$ can be written as $sum from i equals 1 to 5, a sub i.$ We read this as “the sum of a sub i from i equals one to five.” The symbol $∑$ means to add and the i is the index of summation. The 1 tells us where to start (initial value) and the 5 tells us where to end (terminal value).

When we add a finite number of terms, we call the sum a partial sum.

The index does not always have to be i; we can use any letter, but i and k are commonly used. The index does not have to start with 1 either—it can start and end with any nonnegative integer.

Example 2

Expand the partial sum and find its value: $\sum_{k = 0}^{3} \frac{1}{k!} .$

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	$\sum_{k = 0}^{3} \frac{1}{k!}$
We substitute the values 0, 1, 2, 3, in order.	$\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!}$
Evaluate the factorials.	$\frac{1}{1} + \frac{1}{1} + \frac{1}{2} + \frac{1}{6}$
Simplify.	$1 + 1 + \frac{3}{6} + \frac{1}{6}$
Simplify.	$\frac{16}{6}$
Simplify.	$\frac{8}{3}$
	$\sum_{k = 0}^{3} \frac{1}{k!} = \frac{8}{3}$

Use Summation Notation to Write a Sum

In the last two examples, we went from summation notation to writing out the sum. Now we will start with a sum and change it to summation notation. This is very similar to finding the general term of a sequence. We will need to look at the terms and find a pattern. Often the patterns involve multiples or powers.

Example 3

Write the sum using summation notation: $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} .$

Show solution

	$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$
	$n: 1, 2, 3, 4, 5$
We look for a pattern in the terms.	Terms: $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}$
The numerators are all one.	Pattern: $\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \dots, \frac{1}{n}$
The denominators are the counting numbers from one to five.	The sum written in summation notation is
	$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} = \sum_{n = 1}^{5} \frac{1}{n} .$

When the terms of a sum have negative coefficients, we must carefully analyze the pattern of the signs.

Example 4

Write the sum using summation notation: $−1 + 8 - 27 + 64 - 125 .$

Show solution



We look for a pattern in the terms.
The signs of the terms alternate, and the odd terms are negative.
The numbers are the cubes of the counting numbers from one to five.

	The sum written in summation notation is
	$- 1 + 8 - 27 + 64 - 125 = \sum_{n = 1}^{5} {(- 1)}^{n} \cdot n^{3}$

Practice Makes Perfect

Find the Partial Sum

In the following exercises, expand the partial sum and find its value.

$sum from i equals 1 to 5, i squared$

Show solution

$1 + 4 + 9 + 16 + 25 = 55$

$sum from i equals 1 to 5, i cubed$

$\sum_{i = 1}^{6} (2 i + 3)$

Show solution

$5 + 7 + 9 + 11 + 13 + 15 = 60$

$\sum_{i = 1}^{6} (3 i - 2)$

$\sum_{i = 1}^{4} 2^{i}$

Show solution

$2 + 4 + 8 + 16 = 30$

$\sum_{i = 1}^{4} 3^{i}$

$\sum_{k = 0}^{3} \frac{4}{k!}$

Show solution

$\frac{4}{1} + \frac{4}{1} + \frac{4}{2} + \frac{4}{6} = \frac{32}{3} = 10 \frac{2}{3}$

$\sum_{k = 0}^{4} - \frac{1}{k!}$

$\sum_{k = 1}^{5} k (k + 1)$

Show solution

$2 + 6 + 12 + 20 + 30 = 70$

$\sum_{k = 1}^{5} k (2 k - 3)$

$\sum_{n = 1}^{5} \frac{n}{n + 1}$

Show solution

$\frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \frac{4}{5} + \frac{5}{6} = \frac{71}{20}$

$\sum_{n = 1}^{4} \frac{n}{n + 2}$

Use Summation Notation to write a Sum

In the following exercises, write each sum using summation notation.

$\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + \frac{1}{243}$

Show solution

$\sum_{n = 1}^{5} \frac{1}{3^{n}}$

$\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \frac{1}{256}$

$1 + \frac{1}{8} + \frac{1}{27} + \frac{1}{64} + \frac{1}{125}$

Show solution

$\sum_{n = 1}^{5} \frac{1}{n^{3}}$

$\frac{1}{5} + \frac{1}{25} + \frac{1}{125} + \frac{1}{625}$

$2 + 1 + \frac{2}{3} + \frac{1}{2} + \frac{2}{5}$

Show solution

$\sum_{n = 1}^{5} \frac{2}{n}$

$3 + \frac{3}{2} + 1 + \frac{3}{4} + \frac{3}{5} + \frac{1}{2}$

$3 - 6 + 9 - 12 + 15$

Show solution

$\sum_{n = 1}^{5} {(−1)}^{n + 1} 3 n$

$−5 + 10 - 15 + 20 - 25$

$−2 + 4 - 6 + 8 - 10 + \dots + 20$

Show solution

$\sum_{n = 1}^{10} {(−1)}^{n} 2 n$

$1 - 3 + 5 - 7 + 9 + \dots + 21$

$14 + 16 + 18 + 20 + 22 + 24 + 26$

Show solution

$\sum_{n = 1}^{7} (2 n + 12)$

$9 + 11 + 13 + 15 + 17 + 19 + 21$

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	$sum from i equals 1 to 5, 2 i$
We substitute the values 1, 2, 3, 4, 5 in order.	$2 \cdot 1 + 2 \cdot 2 + 2 \cdot 3 + 2 \cdot 4 + 2 \cdot 5$
Simplify.	$2 + 4 + 6 + 8 + 10$
Add.	$30$
	$\sum_{i = 1}^{5} 2 i = 30$