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Sigma notation

Read and write sigma notation, count terms correctly, evaluate arithmetic and geometric sums with shifted bounds, and solve equations and infinite tails.

Before you startArithmetic and geometric series; sums to infinity

01 / Read a sigma expression

Substitute every integer index from the lower limit to the upper limit.

The capital Greek letter sigma, ∑, means “add”. The lower limit supplies the starting index; the upper limit supplies the final index. Include both.

$\sum_{k = 3}^{8} (4k − 1)$ = 11 + 15 + 19 + 23 + 27 + 31 = 126

Here k takes the values 3, 4, 5, 6, 7 and 8. There are 8 − 3 + 1 = 6 terms. The upper limit is not automatically the number of terms.

Choose exactly which terms to addExplore

Top row: the index. Bottom row: the term at that index. Both selected endpoints count.

Indices 2 through 6: 5 terms; sum = 45.

5 + 7 + 9 + 11 + 13 = 45.

Term rule First included index 2Last included index 6

Moving one endpoint past the other brings them together, so at least one term is always selected.

Watch the included indices

Pause, replay or seek freely. The notes explain the same idea and stay in view.

02 / Write a list as a sum

Choose a term rule and check both endpoints.

Write 7 + 12 + 17 + 22 + 27 using sigmaWorked example

The kth term of the list is 5k + 2, starting at k = 1

Check k = 1 gives 7 and k = 5 gives 27.

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

This is one valid representation.

$\sum_{k = 0}^{4} (7 + 5k)$

Starting at zero shifts the term rule. The sum is unchanged.

The index is a dummy variable: renaming k to j everywhere inside the same sum does not change its value. Do not change its limits or term rule accidentally when renaming it.

A quantity independent of the index is constant throughout the sum. For example, summing x five times gives 5x, even if x itself is unknown.

03 / Evaluate arithmetic sums

Find the first term, last term and inclusive count.

Evaluate $\sum_{k = 3}^{8} (4k − 1)$ Worked example

First term = 4(3) − 1 = 11; last term = 4(8) − 1 = 31

Use the actual lower and upper indices.

Number of terms = 8 − 3 + 1 = 6

The interval includes both endpoints.

Sum = 6(11 + 31)/2 = 126

These terms are arithmetic with difference 4.

For a constant c and integers p ≤ q, the sum of c from index p to q is (q − p + 1)c. If p = q, there is exactly one term, not zero.

04 / Separate sums carefully

Addition and constant multiples can be split; products cannot in general.

$\sum_{k = 1}^{n} (3k − 2)$ = 3 $\sum_{k = 1}^{n} k$ − $\sum_{k = 1}^{n} 2$

The constant 2 is added n times. Since 1 + 2 + … + n = n(n + 1)/2, the result is 3n(n + 1)/2 − 2n = n(3n − 1)/2.

Sum rules are linear: Σ(auₖ + bvₖ) = aΣuₖ + bΣvₖ for constants a and b over the same bounds. But Σ(uₖvₖ) generally differs from (Σuₖ)(Σvₖ).

A quick product counterexampleWorked example

1² + 2² = 5

Sum the products term by term.

(1 + 2)(1 + 2) = 9

Multiplying sums introduces extra cross terms.

When adding sums with different limits, first identify which indices overlap. You cannot combine them as though their bounds matched.

05 / Geometric sums with shifted limits

The first included value determines the first term of your sum.

Evaluate $\sum_{k = 2}^{6} 5 × 3ᵏ⁻¹$ Worked example

First included term = 5 × 3¹ = 15

The original sequence’s first term 5 is not included.

Ratio = 3; count = 6 − 2 + 1 = 5

The exponent increases by 1 at each step.

Sum = 15(3⁵ − 1)/(3 − 1) = 1815

Use the selected block as a new geometric series.

Evaluate $\sum_{k = 0}^{5} 3 × 2ᵏ$ Worked example

First term = 3 × 2⁰ = 3; count = 6

Starting at zero still includes a term.

Sum = 3(2⁶ − 1) = 189

The number of terms is 6, even though the upper index is 5.

For powers such as r^(2k), the multiplier when k increases by 1 is r². The common ratio comes from the change between consecutive included terms.

06 / Shift indices or subtract an earlier sum

Preserve the same terms when changing the notation.

Reindex $\sum_{k = 3}^{8} (4k − 1)$ Worked example

Set j = k − 2, so k = j + 2

Translate both the limits and the expression.

k = 3 gives j = 1; k = 8 gives j = 6

The number of terms remains six.

$\sum_{j = 1}^{6} (4j + 7)$

Because 4(j + 2) − 1 = 4j + 7.

Alternatively, subtract early terms:

$\sum_{k = 3}^{8} f(k)$ = $\sum_{k = 1}^{8} f(k)$ − $\sum_{k = 1}^{2} f(k)$

Subtract through index 2, not index 3, to keep the first desired term. Renaming the dummy index alone is different from shifting it: a shift changes the bounds and term expression.

07 / Equations and inequalities

Write the sum as a function of the unknown upper index.

Find a positive integer n with $\sum_{k = 1}^{n} (2k + 3)$ = 140Worked example

Sum = n(5 + 2n + 3)/2 = n(n + 4)

The first term is 5, the last is 2n + 3, and there are n terms.

n² + 4n − 140 = (n − 10)(n + 14) = 0

Solve the quadratic.

n = 10

The negative root −14 is not a valid upper index here.

Least integer n ≥ 4 with $\sum_{k = 4}^{n} (2k + 1)$ ≥ 120Worked example

Sum = (n − 3)(9 + 2n + 1)/2 = (n − 3)(n + 5)

The number of terms is n − 3.

At n = 10 the sum is 105; at n = 11 it is 128

Every new included term is positive.

The least n is 11

Adjacent values and monotonicity establish the first crossing.

Respect any restriction on the upper index. A formula derived for a nonempty sum from 4 to n assumes n ≥ 4.

08 / Infinite tails

Start with the first term that is actually included.

Evaluate $\sum_{k = 4}^{∞} 6(⅓)ᵏ$ Worked example

First included term = 6(⅓)⁴ = 2/27

The lower limit is 4.

Ratio = ⅓, with |r| < 1

The infinite tail converges.

Tail sum = (2/27)/(1 − ⅓) = 1/9

Apply the sum-to-infinity formula to this tail.

Find the least integer n ≥ 3 for which $\sum_{k = 3}^{n} (2k + 1)$ > 80.

Hint

The sum is (n − 2)(n + 4).

Worked solution

At n = 8 the sum is 72; at n = 9 it is 91. All newly added terms are positive, so the least n is 9.

10 / Recap

Sigma notation specifies an index range and a term rule.

Include both limits: p to q contains q − p + 1 terms.
Substitute the actual lower limit to find the first included term.
A constant is added once for each included index.
Shifting the index changes both bounds and the term expression.
For a selected block, subtract only the terms before it.
Check integer bounds and convergence where relevant.

Back to sequences and series →

Section 1 of 10 · Read a sigma expression