Sigma notation stores a sum as two separate things: a rule for the terms and a range for the index. Reading one means expanding it, and writing one means choosing where the index starts, so the same sum has many correct sigmas. The manipulation worth learning is the shift, which moves the limits and the summand together and leaves the value alone.
Core ideas
Nothing new is being computed here. What is new is that the rule for the terms and the range of the index are now written in different places.
Σ says: substitute each integer of the range into the expression beside it, and add the results. Every sigma with whole-number limits could be written out with plus signs instead. It is worth having because it keeps the rule for the terms apart from how many of them there are.
The letter is a placeholder that exists between the limits and nowhere else. Summing 2k - 1 over k from 1 to 5 and summing 2j - 1 over j from 1 to 5 are the same number, 25, written twice. Outside the sigma the letter means nothing, which is why the next sum on the page may reuse it.
From k = 1 to k = n there are n terms, which is the case that hides the rule. From k = 4 to k = 11 there are 8, not 7. Both endpoints are included, so subtracting the limits counts the gaps between terms rather than the terms.
The material
The four parts of the symbol, the three rules it obeys, and the substitution that moves the limits.
For a sequence aₖ and integers m ≤ n, the symbol Σ aₖ with k running from m to n means aₘ + aₘ₊₁ + … + aₙ. The letter k is the index of summation, m is the lower limit, n the upper limit, and the expression being summed is the summand.
Displayed, the limits sit under and over the Σ. In a line of prose there is no room for that, so this course writes the Σ, then the summand, and says the range in words. If the upper limit is below the lower one there are no terms and the sum is 0, which is the number you have when you have added nothing.
The first thing to do with a sigma you have not met is expand it. Substituting every value of the index in turn is slow and it is also the definition, so it can never be wrong.
Two things the notation permits that are easy to miss. The summand need not mention the index, in which case every term is the same and the limits alone decide how many. And the lower limit need not be 1: starting at 0 is common, and starting at 2 is what you have left after peeling the first term off something else.
Σ 7 over k from 1 to 4 is 7 + 7 + 7 + 7 = 28. The k appears nowhere in the summand, and four terms is what the limits asked for. Σ 2ᵏ over k from 0 to 3 is 1 + 2 + 4 + 8 = 15, and the first term is 2⁰ = 1 rather than 2. Σ (-1)ᵏ k over k from 2 to 5 is 2 - 3 + 4 - 5 = -2, where the (-1)ᵏ is doing nothing but setting signs.
Going the other way has two halves. Find a formula that produces the terms from an index, then choose limits that produce exactly the terms you want and no others. The second half is where the choice lies, because a sum has no single correct sigma.
3 + 5 + 7 + 9 + 11 is 35, and all three of these produce it: Σ (2k + 1) over k from 1 to 5; Σ (2k + 3) over k from 0 to 4; Σ (2k - 3) over k from 3 to 7. Each has five terms, each starts at 3 and ends at 11. Which one you write is a matter of where you chose to start counting, and nothing else.
For any sequences aₖ and bₖ, any constant c, and any range of k from m to n: Σ c·aₖ = c·Σ aₖ, and Σ (aₖ + bₖ) = Σ aₖ + Σ bₖ, and Σ c = (n - m + 1)c.
The first two are the distributive law and the freedom to reorder a finite sum, written for many terms at once. The third is the constant summand: one c per value of the index, so the multiplier is the term count and not n.
What is not on that list is a rule for products, and it is the one people reach for. Σ (aₖbₖ) is not (Σ aₖ)(Σ bₖ): take k = 1, 2 with aₖ = bₖ = k, and the left side is 1 + 4 = 5 while the right side is 3 · 3 = 9. Nor does a reciprocal pass through — with terms 1 and 2, Σ (1/aₖ) is 1 + 1/2 = 3/2 and 1/(Σ aₖ) is 1/3. Sigma abbreviates addition, so addition and constant multiples pass through it; a product, a quotient or a power of the terms has to be expanded before anything can be said about it.
Two sums can only be combined term by term when their indices run over the same range, and most closed formulas are written for a sum starting at 1. Both situations call for the same move: change which integers the index runs over without changing which numbers are added.
That is the whole rule: k = j + c sends the limits m and n to m - c and n - c, and the summand f(k) to f(j + c). Both halves move or neither does. Expanding the first term of each form and comparing is the check, and a shift made in the wrong direction misses by exactly the amount shifted — obvious there, invisible afterwards.
Interactive
One sum written four ways: three that are equal and one that only looks it
Method
Four steps, and the first three are done before any adding starts.
Index letter, lower limit, upper limit, summand. Writing them down separately is what makes you notice that the index is absent from the summand, or that the sum does not start at 1.
n - m + 1, before evaluating anything. The count is what you check the expansion against, and a wrong count is an error that leaves every line of the arithmetic looking perfectly normal.
Substitute m, m + 1 and n. A misread summand shows up in the first term, and a misread upper limit shows up in the last, both while they are still cheap to fix.
To move the lower limit, put k = j + c and rewrite the limits and the summand in the same step. Then expand the first term of the new form and confirm it equals the first term of the old one.
One sigma, three jobs: count its terms, add them, and rewrite it to start at k = 1 without changing a single number.
The term count is the line that catches people. 6 - 2 is 4, and there are five terms, because k = 2 and k = 6 are both included. Every wrong answer built on four terms is wrong by one whole term, which for this sum is between 5 and 17.
The shift moved the limits down by 1 and the index inside up by 1. The value survived because k = j + 1 renames the integers fed to the summand without changing which numbers come out. The check did not establish that; it established that this particular shift was carried out correctly.
For a faded re-indexing, use Σ (4k + 1) from k = 3 to k = 7. The supplied count is 7 - 3 + 1 = 5, and the substitution for a range beginning at j = 1 is k = j + 2. Finish the new summand and upper limit, expand both forms, and verify their first term, last term and total independently before opening the quiz.
You should be able to state a sum's term count before evaluating it, write a given list of numbers in sigma notation in at least two ways, and move a lower limit to 1 by substitution. Expanding the first term of both forms is the check, and it costs one line.
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