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Sigma notation

Patterns and sequences · level 7, lesson 1 of 6

A question to start

Here are two long sums:

\[1^2 + 2^2 + 3^2 + 4^2 + \cdots + 99^2 + 100^2\]
\[3 + 5 + 7 + 9 + \cdots + 199 + 201\]

Writing them out in full would take a page each, and the "\(\cdots\)" in the middle asks the reader to guess the pattern. Is the second one the odd numbers? Does it have 100 terms or 99? Mathematicians wanted a way to write such a sum in one short line, with no guessing at all: what you add, where you start and where you stop. The answer is a single Greek letter, and by the end of this lesson you will read and write it as easily as a plus sign.

The idea

The symbol

The Greek capital letter sigma, \(\Sigma\), is the Greek S, for "sum". A sum written with it looks like this:

\[\sum_{k=1}^{5} k^2\]

and you read it as "the sum of \(k^2\), for \(k\) from 1 to 5". It has four parts.

Σ 5 k = 1 k² last value of k what you add the index and its first value
  • The index (here \(k\)) is a counter. It takes whole-number values, one after another.
  • The number under the sigma, \(k = 1\), is the first value (also called the lower limit).
  • The number on top, 5, is the last value (the upper limit). The index goes up by 1 each time until it reaches it.
  • The expression after the sigma, \(k^2\), is the summand: the thing you work out for each value of \(k\) and then add.

So

\[\sum_{k=1}^{5} k^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55.\]

The recipe is always the same: list the values of the index, put each one into the summand, add the results.

The two sums from the start

The first sum is \(\displaystyle\sum_{k=1}^{100} k^2\). Short, and nothing is left to guess.

For the second sum, the odd numbers can be written \(2k + 1\). When \(k = 1\) that is 3, and when \(k = 100\) it is 201. So

\[3 + 5 + 7 + \cdots + 201 = \sum_{k=1}^{100} (2k + 1),\]

and now we can see that it has exactly 100 terms. Answering "how many terms?" is one of the most useful things the notation does for you.

How many terms?

If the index runs from \(m\) to \(n\) (whole numbers, \(m \le n\)), the number of terms is

\[n - m + 1.\]

Why the "+1"? Think of fence posts. From \(k = 3\) to \(k = 7\) the index takes the values 3, 4, 5, 6, 7: that is five values, but \(7 - 3\) is only 4. Subtracting counts the gaps between the values; there is always one more value than gaps.

The name of the index does not matter

\[\sum_{k=1}^{5} k^2, \qquad \sum_{j=1}^{5} j^2, \qquad \sum_{i=1}^{5} i^2\]

all mean \(1 + 4 + 9 + 16 + 25\). The index is a dummy variable: it is used up inside the sum and does not appear in the answer. (The answer 55 has no \(k\) in it.) The letters \(i\), \(j\), \(k\), \(n\) and \(r\) are the usual choices. But be careful when the summand has another letter in it:

\[\sum_{k=1}^{4} (x + k) = (x + 1) + (x + 2) + (x + 3) + (x + 4) = 4x + 10.\]

Here \(x\) is not the index. It stays the same in every term, and it does appear in the answer.

Changing the start (re-indexing)

The same sum can be written with different limits. The powers of 2 from 1 to 16 are

\[1 + 2 + 4 + 8 + 16 = \sum_{k=0}^{4} 2^k = \sum_{j=1}^{5} 2^{\,j-1}.\]

The second form comes from the first by putting \(j = k + 1\). When \(k\) runs 0, 1, 2, 3, 4, the new index \(j\) runs 1, 2, 3, 4, 5, and \(2^k\) becomes \(2^{\,j-1}\). Shift the limits one way and the summand the other way, and the sum is unchanged. This is exactly why the formula for a geometric sequence has \(r^{\,n-1}\) when we count from 1, and \(r^{\,n}\) when we count from 0.

Three rules for working with sums

Every rule below is just ordinary arithmetic (regrouping and the distributive law), written in the new shorthand.

Rule 1: a constant added up. If the summand does not change, you are adding the same number again and again:

\[\sum_{k=1}^{n} c = \underbrace{c + c + \cdots + c}_{n \text{ times}} = nc.\]

So \(\displaystyle\sum_{k=1}^{10} 4 = 40\), not 4. (A sum from \(m\) to \(n\) of a constant \(c\) is \((n - m + 1)c\).)

Rule 2: constants come out. \(\displaystyle\sum_{k=1}^{n} c\,a_k = c\sum_{k=1}^{n} a_k\), because \(ca_1 + ca_2 + \cdots + ca_n = c(a_1 + a_2 + \cdots + a_n)\).

Rule 3: sums split. \(\displaystyle\sum_{k=1}^{n} (a_k + b_k) = \sum_{k=1}^{n} a_k + \sum_{k=1}^{n} b_k\), because you may add the numbers in any order: collect all the \(a\)'s, then all the \(b\)'s.

With these three rules and one known sum, \(\displaystyle\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\) (Gauss's sum from level 5), you can add up any summand of the form \(pk + q\) in one line.

Splitting the range

You can also cut the range of the index in two:

\[\sum_{k=1}^{20} k = \sum_{k=1}^{9} k + \sum_{k=10}^{20} k.\]

Turned round, this gives a handy way to add a sum that does not start at 1: the sum from 10 to 20 is the sum from 1 to 20 minus the sum from 1 to 9. Be careful to subtract up to 9, not up to 10, or you would remove the term 10 that you want to keep.

What sigma cannot do

There is no rule for products. In general

\[\sum_{k=1}^{n} a_k b_k \ne \left(\sum_{k=1}^{n} a_k\right)\left(\sum_{k=1}^{n} b_k\right).\]

Try \(n = 2\) with \(a_k = b_k = k\): the left side is \(1 + 4 = 5\), the right side is \(3 \times 3 = 9\). In the same way, \(\sum k^2\) is not \(\left(\sum k\right)^2\). (In lesson 5 you will find the true formula for \(\sum k^2\), and a surprise about \(\sum k^3\).)

Worked examples

1. Expand and evaluate. Find \(\displaystyle\sum_{k=3}^{7} (2k + 1)\).

  • The index takes the values 3, 4, 5, 6, 7: that is \(7 - 3 + 1 = 5\) terms.
  • The terms are \(7, 9, 11, 13, 15\).
  • Their sum is \(7 + 9 + 11 + 13 + 15 = 55\).

2. Use the rules. Find \(\displaystyle\sum_{k=1}^{10} (3k + 2)\) without listing the terms.

\[\sum_{k=1}^{10} (3k + 2) = 3\sum_{k=1}^{10} k + \sum_{k=1}^{10} 2 = 3 \times \frac{10 \times 11}{2} + 10 \times 2 = 3 \times 55 + 20 = 185.\]

Check by listing: \(5 + 8 + 11 + \cdots + 32\) is an arithmetic sequence with 10 terms, first 5 and last 32, so its sum is \(10 \times \frac{5 + 32}{2} = 185\). ✓

3. Write a sum in sigma notation. Write \(4 + 7 + 10 + \cdots + 61\) with a sigma.

  • The terms go up by 3, so the summand looks like \(3k + \text{something}\).
  • With \(k = 1\) we want 4, so the summand is \(3k + 1\).
  • The last term: \(3k + 1 = 61\) gives \(k = 20\).
\[4 + 7 + 10 + \cdots + 61 = \sum_{k=1}^{20} (3k + 1).\]

(Another correct answer is \(\displaystyle\sum_{k=0}^{19} (3k + 4)\). Both have 20 terms.)

4. A sum that does not start at 1. Find \(\displaystyle\sum_{k=10}^{20} k\).

\[\sum_{k=10}^{20} k = \sum_{k=1}^{20} k - \sum_{k=1}^{9} k = \frac{20 \times 21}{2} - \frac{9 \times 10}{2} = 210 - 45 = 165.\]

Check: there are \(20 - 10 + 1 = 11\) terms, from 10 to 20, so the sum is \(11 \times \frac{10 + 20}{2} = 165\). ✓

5. A sum that collapses. Find \(\displaystyle\sum_{k=1}^{99} \left(\frac{1}{k} - \frac{1}{k+1}\right)\).

Write out the first few terms and the last:

\[\left(1 - \tfrac12\right) + \left(\tfrac12 - \tfrac13\right) + \left(\tfrac13 - \tfrac14\right) + \cdots + \left(\tfrac{1}{99} - \tfrac{1}{100}\right).\]

Every \(-\frac12\) is cancelled by the next \(+\frac12\), every \(-\frac13\) by the next \(+\frac13\), and so on. Only the very first and very last pieces survive:

\[1 - \frac{1}{100} = \frac{99}{100}.\]

A sum like this, where neighbouring terms cancel, is called telescoping, after the old folding telescope that slides shut. Since \(\frac{1}{k} - \frac{1}{k+1} = \frac{1}{k(k+1)}\), this also tells you that \(\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \cdots + \frac{1}{99 \times 100} = \frac{99}{100}\). You will use the telescoping idea again in lesson 5.

Try it

  1. Expand and evaluate \(\displaystyle\sum_{k=1}^{4} (5 - k)\).
  2. How many terms does \(\displaystyle\sum_{k=7}^{30} k^3\) have?
  3. Write in sigma notation: (a) \(2 + 4 + 6 + \cdots + 50\); (b) \(1 + \tfrac12 + \tfrac14 + \cdots + \tfrac{1}{64}\).
  4. Find \(\displaystyle\sum_{k=1}^{50} 6\).
  5. Use the rules to find \(\displaystyle\sum_{k=1}^{20} (4k - 1)\).
  6. Find \(\displaystyle\sum_{k=21}^{40} k\).
  7. Rewrite \(\displaystyle\sum_{k=3}^{12} (k - 2)^2\) so that the index starts at 1, and then evaluate it, using \(1^2 + 2^2 + \cdots + 10^2 = 385\).
  8. Show that \(\displaystyle\sum_{k=1}^{n} \left(\sqrt{k+1} - \sqrt{k}\right) = \sqrt{n+1} - 1\), and use it to find the sum when \(n = 99\).
Answers
  1. The terms are \(4, 3, 2, 1\), so the sum is 10.
  2. \(30 - 7 + 1 =\) 24 terms.
  3. (a) The terms are \(2k\) for \(k = 1\) to 25: \(\displaystyle\sum_{k=1}^{25} 2k\). (b) The terms are powers of \(\tfrac12\) from \(\left(\tfrac12\right)^0\) to \(\left(\tfrac12\right)^6\): \(\displaystyle\sum_{k=0}^{6} \left(\tfrac12\right)^k\) (or \(\displaystyle\sum_{k=1}^{7} \left(\tfrac12\right)^{k-1}\)).
  4. A constant added 50 times: \(50 \times 6 =\) 300.
  5. \(4 \times \frac{20 \times 21}{2} - 20 \times 1 = 4 \times 210 - 20 =\) 820. (Check: first term 3, last 79, 20 terms, \(20 \times \frac{3 + 79}{2} = 820\).)
  6. \(\frac{40 \times 41}{2} - \frac{20 \times 21}{2} = 820 - 210 =\) 610.
  7. Put \(j = k - 2\). When \(k = 3\), \(j = 1\); when \(k = 12\), \(j = 10\). The sum is \(\displaystyle\sum_{j=1}^{10} j^2 =\) 385.
  8. Writing out the terms, \((\sqrt2 - \sqrt1) + (\sqrt3 - \sqrt2) + \cdots + (\sqrt{n+1} - \sqrt{n})\): each \(+\sqrt{k}\) cancels the \(-\sqrt{k}\) in the next term, leaving \(\sqrt{n+1} - \sqrt1 = \sqrt{n+1} - 1\). With \(n = 99\) the sum is \(\sqrt{100} - 1 =\) 9.

Common mistake

Treating a constant as a single term. \(\displaystyle\sum_{k=1}^{10} 4\) is \(4 + 4 + \cdots + 4 = 40\), not 4. The same slip happens inside bigger sums: in \(\displaystyle\sum_{k=1}^{10} (3k + 2)\) the 2 is added ten times, giving 20.

Off-by-one counting. The sum from \(k = 10\) to \(k = 20\) has 11 terms, not 10. And when you subtract to get it, you take away the sum up to 9, not up to 10.

Where it leads