How to Write a Sum in Sigma Notation

To write in sigma notation find a pattern in the expanded terms, write a general term that produces every term, then pick an index start and an upper bound that matches the number of terms. Every expanded series has at least one correct Σ form, and often more than one because you can shift the index. The most common error is writing a general term that fails for the first term or the last term, so test at both ends.

This is the reverse of evaluating a sum: instead of plugging an index range into a formula, you look at a list of numbers and reconstruct the formula that generates them. The method works for arithmetic, geometric, fractional, and alternating series, and it is the skill precalculus textbooks like OpenStax Precalculus 2e section 9.4 (Series and Their Notations) expect before you move to infinite series or Riemann sums.

Look at the expanded terms and identify how each term relates to its position. Ask three questions:

• Is there a constant difference between consecutive terms? That is an arithmetic series.
• Is there a constant ratio between consecutive terms? That is a geometric series.
• Do the terms look like squares, cubes, or reciprocals of something? That is a power pattern.

Write the terms in a row and number their positions starting at 1. For the series 5 + 9 + 13 + 17 + 21, the difference between each term is 4, so the pattern is arithmetic with common difference 4. For 3 + 6 + 12 + 24 + 48, the ratio is 2, so the pattern is geometric with common ratio 2. For 1 + 4 + 9 + 16 + 25, each term is a square: 1², 2², 3², 4², 5².

If the signs alternate, for example 1 − 3 + 5 − 7 + 9, you have an alternating series. The sign pattern uses a factor of (−1)^k or (−1)^(k+1) depending on whether the first term is positive or negative.

The general term is a formula in the index variable k that produces the k-th term when you substitute k = 1, 2, 3, and so on. For the arithmetic series 5 + 9 + 13 + 17 + 21, the first term is 5 and the common difference is 4. The arithmetic general term is aₖ = a₁ + (k − 1)d = 5 + (k − 1)×4 = 4k + 1. Check: k=1 gives 5, k=2 gives 9, k=3 gives 13, k=4 gives 17, k=5 gives 21. That works.

For a geometric series like 3 + 6 + 12 + 24 + 48, the first term is 3 and the common ratio is 2. The geometric general term is aₖ = a₁·r^(k−1) = 3·2^(k−1). Check: k=1 gives 3, k=2 gives 6, k=3 gives 12, k=4 gives 24, k=5 gives 48. That works.

For squares, the general term is simply k². For fractions such as 1/2 + 1/4 + 1/8 + 1/16, the general term is 1/2^k. For the series 1/1 + 1/4 + 1/9 + 1/16 + 1/25, the general term is 1/k².

Once you have the general term, decide where the index starts. Most textbook problems start at k = 1 or k = 0. The lower bound is the smallest value of k that gives the first term. The upper bound is the largest value of k that gives the last term. For a series with n terms, if you start at k = 1, the upper bound is n. If you start at k = 0, the upper bound is n − 1.

For the arithmetic series 5 + 9 + 13 + 17 + 21, we have 5 terms. Starting at k = 1 with general term 4k + 1, the upper bound is 5. So the sigma notation is Σ_{k=1}^{5} (4k + 1). Starting at k = 0 with general term 4(k + 1) + 1 = 4k + 5, the upper bound is 4, giving Σ_{k=0}^{4} (4k + 5). Both are correct, and both sum to 65.

For the geometric series 3 + 6 + 12 + 24 + 48, starting at k = 1 gives Σ_{k=1}^{5} 3·2^(k−1). Starting at k = 0 gives Σ_{k=0}^{4} 3·2^k. Both produce the same 5 terms. This is called an index shift, and it is one reason sigma notation can have more than one correct answer.

When a series alternates between positive and negative terms, you need a factor of (−1) raised to a power. If the first term is positive, use (−1)^(k+1) when starting at k = 1, because (−1)^(1+1) = (−1)² = +1. If the first term is negative, use (−1)^k when starting at k = 1, because (−1)^1 = −1.

For the series 1 − 3 + 5 − 7 + 9, the terms are positive, negative, positive, negative, positive. The absolute values are odd numbers: 1, 3, 5, 7, 9. The general term for the absolute values is (2k − 1). To get the alternating sign starting positive, use (−1)^(k+1). So the sigma notation is Σ_{k=1}^{5} (−1)^(k+1) (2k − 1). Testing: k=1 gives (+1)(1) = 1, k=2 gives (−1)(3) = −3, k=3 gives (+1)(5) = 5, k=4 gives (−1)(7) = −7, k=5 gives (+1)(9) = 9. That works.

If the first term were negative, for example −1 + 3 − 5 + 7 − 9, you would use (−1)^k, giving Σ_{k=1}^{5} (−1)^k (2k − 1).

Arithmetic: 2 + 7 + 12 + 17 + 22

Common difference is 5, first term is 2. General term: aₖ = 2 + (k − 1)·5 = 5k − 3.

Geometric: 4 + 12 + 36 + 108 + 324

Common ratio is 3, first term is 4. General term: aₖ = 4·3^(k−1). Five terms, starting at k = 1, upper bound 5. Answer: Σ_{k=1}^{5} 4·3^(k−1).

Squares: 1 + 4 + 9 + 16 + 25 + 36

Each term is k². Six terms, starting at k = 1, upper bound 6. Answer: Σ_{k=1}^{6} k².

Fractions: 1/3 + 1/9 + 1/27 + 1/81

Each term is 1/3^k. Four terms, starting at k = 1, upper bound 4. Answer: Σ_{k=1}^{4} 1/3^k. You could also start at k = 0 with 1/3^(k+1), but the common convention uses k = 1.

Alternating: 2 − 4 + 6 − 8 + 10 − 12

Absolute values are even numbers: 2, 4, 6, 8, 10, 12. General term for absolute values: 2k. First term is positive, so use (−1)^(k+1). Six terms, starting at k = 1. Answer: Σ_{k=1}^{6} (−1)^(k+1) · 2k.

Sigma notation is not unique. You can shift the index by 1, 2, or any integer as long as you adjust the bounds and the general term accordingly. This is called re-indexing. For the series 3 + 6 + 9 + 12 + 15, you can write:

• Σ_{k=1}^{5} 3k
• Σ_{k=0}^{4} 3(k + 1)
• Σ_{k=2}^{6} 3(k − 1)

All three produce the same five terms. The index variable itself is a dummy variable, the letter does not matter. Σ_{i=1}^{5} 3i is identical to Σ_{k=1}^{5} 3k. The only rule is that the lower bound, upper bound, and the expression inside must agree. When you check your work, try a small-n case: compute the sum for n = 2 or n = 3 by hand and compare to the expanded form.

Convert each expanded series into sigma notation. Write the general term, the index start, and the upper bound. Answers follow the list.

1. 5 + 8 + 11 + 14 + 17
2. 2 + 10 + 50 + 250 + 1250
3. 1 + 8 + 27 + 64 + 125 + 216
4. −1 + 2 − 3 + 4 − 5 + 6
5. 1/2 + 1/4 + 1/8 + 1/16 + 1/32
6. 1 + 1/9 + 1/25 + 1/49

Answers:

1. Σ_{k=1}^{5} (3k + 2)  , arithmetic, d = 3, first term 5.
2. Σ_{k=1}^{5} 2·5^(k−1)  , geometric, r = 5, first term 2.
3. Σ_{k=1}^{6} k³  , cubes of the first 6 integers.
4. Σ_{k=1}^{6} (−1)^k·k  , first term negative, absolute values k.
5. Σ_{k=1}^{5} 1/2^k  , fractions, denominator powers of 2.
6. Σ_{k=1}^{4} 1/(2k−1)²  , squares of odd numbers: 1, 3, 5, 7.

The most frequent mistake when converting series to sigma notation is an off-by-one error in the upper bound. For the series 1 + 3 + 5 + 7 + 9, writing Σ_{k=1}^{5} (2k − 1) works, but writing Σ_{k=1}^{4} (2k − 1) gives only four terms and misses the last one. Count the number of terms and set the upper bound to match.

A second failure case is misapplying the geometric formula when r = 1. For a series of constant terms like 5 + 5 + 5 + 5, the general term is 5, not 5·1^(k−1). The geometric formula for Σ a·r^(k−1) works for r ≠ 1; when r = 1, the sum is simply n·a.

A third error is confusing Σk² with a power series like Σaₙxⁿ. Sum of powers Σk² is a finite polynomial in n, its closed form is n(n+1)(2n+1)/6. A power series is an infinite function of x. The two have nothing to do with each other beyond the word 'power'.

The general term of a sequence is the engine of sigma notation. Once you have the general term, you plug in the index to produce each term of the sum. For sequences that are neither arithmetic nor geometric, for example 1/2 + 2/3 + 3/4 + 4/5, the general term is k/(k+1). That gives Σ_{k=1}^{4} k/(k+1).

For series that involve reciprocals of squares, like 1/1 + 1/4 + 1/9 + 1/16, the general term is 1/k². For cubes of reciprocals, 1/8 + 1/27 + 1/64 + 1/125, the general term is 1/k³ starting at k = 2, giving Σ_{k=2}^{5} 1/k³. Finding the general term is the step that requires the most attention; the index bounds are mechanical once the general term is correct.