Sigma notation

When you reach for it

Adding up the terms of a sequence is one of the most common things done with it. Writing the sum out in full stops being practical once there are more than a handful of terms, and the usual shortcut, a few terms and then dots,

leaves the reader to guess the rule. Is the next term after 8 really 11? Where exactly does the sum stop? Dots show a pattern; they do not state one.

Sigma notation states it. It names the rule that makes each term and the exact range of positions to use, so there is nothing left to guess. Reach for it whenever you need to write a sum of many terms that follow one rule: the total of a sequence’s first n terms, the area of a stack of thin rectangles, a total that builds up week by week. It is also the language of everything that comes after this lesson about series, so reading it fluently matters as much as writing it.

This lesson uses the explicit formulas of arithmetic and geometric sequences, and the language of terms and positions from Sequences, terms and notation.

The procedure

Reading the notation

Here is a sum in sigma notation:

It is read “the sum of 3k - 1, for k from 1 to 6”. The large symbol \Sigma is the capital Greek letter sigma, the Greek S, standing for “sum”. The expression has four parts.

• The index is the letter written under the sigma, here k. It counts the positions.
• The lower limit is the number the index starts at, written after the equals sign under the sigma, here 1.
• The upper limit is the number the index stops at, written above the sigma, here 6.
• The summand is the expression to the right of the sigma, here 3k - 1. It is the rule for the term in position k, exactly an explicit formula a_k = 3k - 1.

The index takes every integer from the lower limit to the upper limit, one at a time, and each value goes into the rule. So k is 1, 2, 3, 4, 5, 6 and the terms are

Then the terms are added. Writing them out with plus signs is called expanding the sum:

The number of terms is the number of integers from the lower limit to the upper limit, counting both ends:

The +1 is there because both limits are used. From 1 to 6 there are 6 - 1 + 1 = 6 terms; from 3 to 7 there are 7 - 3 + 1 = 5, namely 3, 4, 5, 6, 7.

Brackets around the summand matter. In \sum_{k=1}^{6} (3k - 1) the whole of 3k - 1 is the rule. Without brackets, \sum_{k=1}^{6} 3k - 1 would mean \left(\sum_{k=1}^{6} 3k\right) - 1, the sum of 3k and then a single 1 taken away, which is 63 - 1 = 62. When the summand has more than one term, bracket it.

The index is a placeholder

The index only counts. Once the sum is expanded it has disappeared, so the letter chosen for it makes no difference to the value:

Both say “take the rule at positions 1 to 6 and add”. The common choices are i, j, k and n. What does change the value is the limits and the rule. A sum whose upper limit is a letter, such as \sum_{k=1}^{n} (3k - 1), has a value that depends on n, because n says how many terms there are; it does not depend on k.

Writing a sum in sigma notation

To go the other way, from a written-out sum to sigma notation, two things are needed: a rule for the term in position k, and the positions of the first and last terms.

The rule is an explicit formula, found just as for a sequence. For 5 + 9 + 13 + \cdots + 41 the terms rise by 4 each time, so the sequence is arithmetic with first term 5 and common difference 4, and the term in position k is

Starting the index at 1 makes the first term a_1 = 5. The upper limit is the position of the last term, 41, found by solving

So

Always check the result by putting the two limits into the rule: k = 1 gives 5 and k = 10 gives 41, the first and last terms.

Shifting the index

The same sum can be written with a different starting value, as long as the rule changes to match. Replace k by k + 1 everywhere in the summand and lower both limits by 1:

The terms are the same list, 2, 5, 8, 11, 14, 17, because the new index meets each term one step earlier: the old position k = 1 is the new position k = 0. A start at 0 is natural for a geometric sum, where r^0 = 1 makes the first term the plain first term.

Properties of sums

Four facts let you work with a sum without expanding it. In each, a_k and b_k are any two rules, c is a number that does not depend on k, and the sums run over the same positions.

A constant factor comes out.

Expanded, the left side is c a_1 + c a_2 + \cdots + c a_n. Every term has the factor c, and the The distributive property takes it out: c(a_1 + a_2 + \cdots + a_n).

A sum or difference splits.

The left side is

The commutative property lets the terms be put in any order, and the associative property lets them be grouped in any way, so all the a terms can be gathered first and the b terms after. A difference splits the same way, since subtracting b_k is adding -b_k and the constant factor -1 comes out.

A constant added n times is n times the constant.

The summand has no k in it, so every one of the n terms is c, and n copies of c add to nc. The sum is not just c: the limits still say there are n terms.

A sum splits at any position. For an integer m with 1 \le m < n,

The first sum on the right takes positions 1 to m, the second positions m + 1 to n, and between them every position is used once. By the associative property, adding the first group and then the second gives the same total as adding everything in order.

Here are all four at work on the opening sum. Split the summand, take the factor 3 out, and count the constant 1 six times:

which matches the expanded sum. Finding a short formula for \sum_{k=1}^{n} k itself, so that no sum ever needs adding term by term, is the subject of Arithmetic series and partial sums.

Worked examples

Boundaries

• The index moves in steps of 1 only. \sum_{k=1}^{6} means k = 1, 2, 3, 4, 5, 6 and nothing in between; k = 1.5 is never used. The limits are integers, and the lower limit is the first value used, not a value to skip.
• The upper limit is normally at least the lower limit. If it is smaller, as in \sum_{k=5}^{2}, no integer lies in the range and there are no terms. Such a sum is taken to be 0, the empty sum, but this lesson never needs one.
• Only a true constant comes out. The factor in \sum c\,a_k = c\sum a_k must not depend on the index. \sum_{k=1}^{3} k \cdot k is 1 + 4 + 9 = 14, but pulling k out in front would be meaningless, since outside the sum there is no k.
• There is no such rule for products or squares. A sum of products is not the product of the sums, and a sum of squares is not the square of the sum. For instance \sum_{k=1}^{2} k^2 = 1 + 4 = 5, while \left(\sum_{k=1}^{2} k\right)^2 = 3^2 = 9.
• The splitting rules need matching positions. \sum (a_k + b_k) = \sum a_k + \sum b_k holds when all three sums run over the same positions; sums over different ranges cannot be merged this way.
• An infinite upper limit means something new. In \sum_{k=1}^{\infty} there is no last term to reach, so the sum cannot be found by adding. It is defined by a limit, and only some such sums have a value; see Infinite geometric series.

Where this goes next

With the notation in hand, the next step is to stop adding term by term. Arithmetic series and partial sums finds a formula for the sum of the first n terms of an arithmetic sequence, so that \sum_{k=1}^{100} (3k - 1) costs one calculation instead of a hundred. Geometric series does the same for geometric sequences, where the sums describe loan repayments and repeated deposits, and Infinite geometric series gives meaning to a sum with infinitely many terms. In calculus, the rectangles of Example 7 grow into sums that approximate an area and then into the definite integral, whose very symbol, a stretched S, was chosen to echo the sigma of a sum.

Exercises

Expanding and evaluating

1. Write out \sum_{k=1}^{5} (4k - 3) and find its value.
2. Find \sum_{n=1}^{4} n^3.
3. Find \sum_{i=2}^{6} (i^2 - i).
4. Find \sum_{k=0}^{5} (-2)^k.
5. Find \displaystyle \sum_{k=1}^{4} \frac{1}{k} as a single fraction.
6. Find \sum_{j=1}^{8} 5.
7. How many terms does \displaystyle \sum_{k=4}^{19} \sqrt{k} have? What are its first and last terms?

Writing a sum in sigma notation

8. Write 4 + 8 + 12 + \cdots + 60 in sigma notation.
9. Write 3 + 9 + 19 + 33 + 51 + 73 in sigma notation. (Compare each term with 2k^2.)
10. Write 81 + 27 + 9 + 3 + 1 in sigma notation.
11. Write -1 + 2 - 3 + 4 - 5 + 6 - 7 in sigma notation and find its value.
12. Write \displaystyle \frac{1}{3} + \frac{2}{4} + \frac{3}{5} + \cdots + \frac{8}{10} in sigma notation.
13. Rewrite \sum_{k=1}^{8} (6k - 1) so that the index starts at 0.

Properties of sums

For exercises 14 to 16, \sum_{k=1}^{10} a_k = 32 and \sum_{k=1}^{10} b_k = -6.

14. Find \sum_{k=1}^{10} 5a_k.
15. Find \sum_{k=1}^{10} (a_k - b_k).
16. Find \sum_{k=1}^{10} (a_k + 4).
17. If \sum_{k=1}^{15} d_k = 120 and \sum_{k=1}^{9} d_k = 50, find \sum_{k=10}^{15} d_k.
18. Given \sum_{k=1}^{10} k = 55 and \sum_{k=1}^{10} k^2 = 385, find \sum_{k=1}^{10} (k^2 - 4k + 2) without expanding.

In a setting

19. A walking plan has the walk on day k of one week be \displaystyle 2 + \frac{k}{2} kilometers, for k = 1 to 7. Write the week’s total distance in sigma notation and find it.
20. Four rectangles, each 1 wide, stand on the x-axis from x = 1 to x = 5. Each one’s height is the value of \displaystyle y = \frac{60}{x} at its left edge. Write their total area in sigma notation and find it.

Reasoning

21. Explain why \sum_{k=1}^{5} k^2 and \sum_{j=1}^{5} j^2 are equal, but \sum_{k=1}^{5} k^2 and \sum_{k=0}^{4} k^2 are not. Find all three.
22. Show with a small example that \sum_{k=1}^{n} a_k b_k is not always equal to \left(\sum_{k=1}^{n} a_k\right)\left(\sum_{k=1}^{n} b_k\right).
23. Prove that \sum_{k=1}^{n} (a_k - b_k) = \sum_{k=1}^{n} a_k - \sum_{k=1}^{n} b_k, naming the property used at each step.
24. A classmate says \sum_{k=1}^{n} 7 = 7, because “the summand is just 7”. Explain what is wrong and give the correct value.
25. Find \displaystyle \sum_{k=1}^{99} \left(\frac{1}{k} - \frac{1}{k + 1}\right) by writing out the first few and the last few terms and explaining why almost all of them cancel.