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Sigma Notation for Finite Sums

Learn to translate between finite sums and sigma notation by identifying the index bounds and substituting each index value into the summand.

Beginner 9 min read

What You'll Learn

  • Read the sigma symbol, summand, index, and inclusive lower and upper bounds.
  • Translate an expanded finite sum into sigma notation.
  • Expand sigma notation by substituting each integer index value.
  • Write sums of constants and powers with the correct index limits.
  • Evaluate finite sums of indexed terms when the required values are given.
  • Recognize and correct errors caused by omitting an endpoint or misreading the index.

Why Use Sigma Notation?

A finite sum can be written by listing every term, but long lists are difficult to read and easy to miscopy. Sigma notation gives a compact way to describe the terms and the precise range of indices that appear.

For example, the expression \(4+7+10+13\) has four terms. These terms follow the rule \(3k+1\) as \(k\) takes the integer values from \(1\) through \(4\). Sigma notation records both that rule and that range:

$$ \sum_{k=1}^{4}(3k+1). $$

The notation does not change which numbers are being added. It makes the pattern and the endpoints explicit.

Definition: For a function or expression \(a_k\) defined at each integer \(k\) from \(m\) through \(n\), the finite sum \(\sum_{k=m}^{n}a_k\) means to add the values \(a_m,a_{m+1},\ldots,a_n\). The lower index \(m\) and upper index \(n\) are both included.

The Parts of a Finite Sum

In \(\sum_{k=m}^{n}a_k\), the large Greek letter \(\sum\) is the summation symbol. The letter \(k\) is the index: it takes the consecutive integer values specified by the bounds. The expression \(a_k\) is the summand, meaning the term to be evaluated and added for each index value. The lower index appears below the sigma, and the upper index appears above it.

$$ \sum_{k=m}^{n}a_k = a_m+a_{m+1}+\cdots+a_n. $$

The index is a placeholder whose values vary across the sum. It is not an additional quantity being added. In the expression \(\sum_{k=2}^{5}(3k-3)\), for instance, the index takes the values \(2,3,4,5\). The summand \(3k-3\) is evaluated separately at each of those values.

The index letter itself is not important. When the summand and bounds are changed consistently, the same sum can be written with a different index letter:

$$ \sum_{k=1}^{3} k^2 = \sum_{j=1}^{3} j^2 = 1^2+2^2+3^2. $$

Here \(k\) and \(j\) play the same role. The index is often called a dummy index because its name has no effect on the value of the sum. The index letter must, however, be used consistently in the summand and its bounds.

Worked Example: Expand and Evaluate an Indexed Expression

Worked Example: Expand and Evaluate an Indexed Expression

Expand \(\sum_{k=2}^{5}(3k-3)\), then evaluate it.

The lower index is \(2\), and the upper index is \(5\). Both are included, so substitute \(k=2,3,4,5\) into the summand \(3k-3\):

$$ \begin{aligned} \sum_{k=2}^{5}(3k-3) &=(3(2)-3)+(3(3)-3)+(3(4)-3)+(3(5)-3)\\ &=(6-3)+(9-3)+(12-3)+(15-3)\\ &=3+6+9+12\\ &=30. \end{aligned} $$

Check the expansion against the bounds: there are \(5-2+1=4\) index values, and the expanded expression has four terms. Substituting each of those values into \(3k-3\) gives \(3,6,9,12\), whose sum is \(30\). Thus the expansion and evaluation agree.

Sums of Constants and Powers

The summand does not have to depend meaningfully on the index. A constant can be written as the summand, in which case that same constant is added once for every index value. If there are \(n-m+1\) integer indices from \(m\) through \(n\), a constant \(c\) appears \(n-m+1\) times.

$$ \sum_{k=m}^{n}c = \underbrace{c+c+\cdots+c}_{n-m+1\text{ terms}} \qquad (m\le n). $$

Powers can also be used as summands. For example, \(\sum_{j=1}^{4}j^2\) means to square each index value and then add the results. The bounds apply to the index \(j\); the exponent \(2\) is part of the summand.

Worked Example: Sum a Constant Over Several Indices

Expand and evaluate \(\sum_{r=3}^{6}5\).

The index values are \(3,4,5,6\). The summand is the constant \(5\), so it contributes once at each of the four indices:

$$ \begin{aligned} \sum_{r=3}^{6}5 &=5+5+5+5\\ &=10+5+5\\ &=15+5\\ &=20. \end{aligned} $$

Check the number of terms: \(6-3+1=4\), matching the four copies of \(5\). Their total is \(4\cdot5=20\), which confirms the expansion.

Worked Example: Expand a Sum of Powers

Expand and evaluate \(\sum_{j=1}^{4}j^2\).

The index \(j\) takes the values \(1,2,3,4\). For each value, calculate the square specified by the summand \(j^2\):

$$ \begin{aligned} \sum_{j=1}^{4}j^2 &=1^2+2^2+3^2+4^2\\ &=1+4+9+16\\ &=5+9+16\\ &=14+16\\ &=30. \end{aligned} $$

Check the endpoints and term count: the indices run from \(1\) to \(4\), inclusive, so there are \(4-1+1=4\) terms. The displayed expansion contains exactly those four squares, and their total is \(30\).

Reading and Writing Sigma Notation

To expand a sigma expression reliably, first identify its index and bounds. Then list every integer index value in order, substitute each value into the entire summand, and add the resulting terms. Keeping the substitution visible helps distinguish the index from the value of the summand.

1
Read the bounds.
Record the integer values beginning at the lower index and ending at the upper index, including both endpoints.
2
Identify the summand.
Keep the complete expression to the right of the sigma together. For example, in \(\sum_{i=1}^{3}(2i+1)\), the summand is \(2i+1\).
3
Substitute each index value.
Evaluate the summand once at each index value, without skipping any values.
4
Add the resulting terms.
After expansion, ordinary arithmetic gives the value of the finite sum.

To write a listed sum in sigma notation, look for a rule that produces each term and choose an index range that visits precisely those terms. For example, the terms \(4,7,10,13\) increase by \(3\). The rule \(3k+1\) produces them for \(k=1,2,3,4\):

$$ \begin{aligned} 3(1)+1&=4,\\ 3(2)+1&=7,\\ 3(3)+1&=10,\\ 3(4)+1&=13. \end{aligned} $$

Therefore the sum is \(\sum_{k=1}^{4}(3k+1)\). A formula for the summand and correct bounds are both needed: the formula determines what each term is, while the bounds determine how many terms appear and which index values generate them.

Worked Example: Add Values Defined by an Index

Suppose \(a_1=2\), \(a_2=5\), and \(a_3=4\). Expand and evaluate \(\sum_{i=1}^{3}(2a_i-1)\).

The index values are \(1,2,3\). At each value, use the corresponding given value of \(a_i\) in the summand:

$$ \begin{aligned} \sum_{i=1}^{3}(2a_i-1) &=(2a_1-1)+(2a_2-1)+(2a_3-1)\\ &=(2(2)-1)+(2(5)-1)+(2(4)-1)\\ &=(4-1)+(10-1)+(8-1)\\ &=3+9+7\\ &=19. \end{aligned} $$

Check the expansion: the bounds require three terms, one for each of \(a_1,a_2,a_3\), and all three appear. Substitution gives \(3,9,7\), whose total is \(19\). This agrees with the stated index values and summand.

A Justification of the Expansion Rule

Expanding a finite sum is not an extra calculation rule imposed on the sigma symbol. It follows from what the notation means: the expression tells us to add one evaluated summand for each permitted index. This can be formalized recursively.

Definition: For \(m\le n\), define the finite sum recursively by \(\sum_{k=m}^{m}a_k=a_m\), and, for each integer \(r\) with \(m<r\le n\), define \(\sum_{k=m}^{r}a_k=\sum_{k=m}^{r-1}a_k+a_r\).

The first rule says that a sum with one permitted index consists of its one term. The second says that increasing the upper bound by one adds exactly the new endpoint term. Together, they justify listing all the terms in order.

1
Start at the lower bound.
When the upper bound is \(m\), the definition gives \(\sum_{k=m}^{m}a_k=a_m\), so the expansion contains exactly the first term.
2
Add the next index.
For any \(r\) with \(m<r\le n\), the recursive definition gives \(\sum_{k=m}^{r}a_k=\sum_{k=m}^{r-1}a_k+a_r\). If the sum through \(r-1\) contains \(a_m+\cdots+a_{r-1}\), the sum through \(r\) contains \(a_m+\cdots+a_{r-1}+a_r\).
3
Reach the upper bound.
Repeating this step for each integer index through \(n\) yields \(\sum_{k=m}^{n}a_k=a_m+a_{m+1}+\cdots+a_n\). Thus every index in the bounds contributes once, and no other term is included.

This justification also explains why the endpoints matter. Omitting the lower or upper endpoint changes the set of terms; using a different upper bound changes how many times the recursive addition is performed.

Common Mistake: Treating an Endpoint as Optional

A frequent error is to use the lower bound as a starting point for counting rather than as an index value to substitute. Consider \(\sum_{i=2}^{4}(i+1)\). A mistaken expansion might be:

$$ \text{Incorrect:}\qquad \sum_{i=2}^{4}(i+1)=2+3+4=9. $$

This calculation has used the index values \(1,2,3\) in the summand, even though the stated bounds are \(2\) through \(4\). The lower index is included, and the summand is \(i+1\), not \(i\). The correct expansion is:

$$ \begin{aligned} \sum_{i=2}^{4}(i+1) &=(2+1)+(3+1)+(4+1)\\ &=3+4+5\\ &=12. \end{aligned} $$

There are \(4-2+1=3\) terms, as required, and substituting \(i=2,3,4\) produces \(3,4,5\). The incorrect result came from confusing the index values with the values of the summand and shifting the range.

Conventions and a Useful Check

In the examples here, the lower bound does not exceed the upper bound. Under the standard convention, if integer bounds are written with the lower bound greater than the upper bound, the sum has no terms and its value is \(0\). For instance, \(\sum_{k=4}^{3}a_k=0\). This convention makes some general formulas convenient, but for ordinary expansions the main task is to ensure that the stated range includes exactly the intended indices.

A useful check is to count the terms. When \(m\le n\) are integer bounds, the index values are \(m,m+1,\ldots,n\), so their number is \(n-m+1\). Compare this count with the number of terms in your expanded expression. This catches many endpoint errors, although you should still check that each individual term comes from the correct substitution.

Key takeaway: A finite sigma expression specifies a summand and an inclusive range of integer indices. Expand it by substituting every index value in that range into the summand, then add the resulting terms.

Check Your Understanding

For each question, pay attention to the index values and include both stated endpoints when expanding.

  1. Identify the index, summand, lower index, and upper index in \(\sum_{r=3}^{7}(2r+1)\).
  2. Expand and evaluate \(\sum_{k=2}^{4}k\).
  3. Write \(6+10+14+18\) in sigma notation using a summand that depends linearly on an index beginning at \(1\).
  4. Expand \(\sum_{j=1}^{3}4\). How many terms appear, and what is the sum?
  5. If \(b_1=3\), \(b_2=1\), and \(b_3=6\), expand and evaluate \(\sum_{i=1}^{3}(b_i+2)\).

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