2.6 Sigma Notation
SL 1.2 SL 1.3
- how to read and write a sum in sigma notation \(\displaystyle\sum\)
- how to evaluate a sum written in sigma notation, by hand and on the GDC
- the rules for working with sigma notation
- how to write a given series in sigma notation
- how arithmetic and geometric series look in sigma notation
Reading sigma notation
Writing out a long sum is slow. Sigma notation describes a whole series in one short expression:
\[ \sum_{r=1}^{n} u_r = u_1 + u_2 + u_3 + \cdots + u_n . \]
- \(\Sigma\) is the Greek capital letter sigma (S for “sum”). It means “add up”.
- \(u_r\) is the general term: a formula for each term.
- \(r\) is the index (a counter). Any letter can be used: \(r\), \(k\), \(i\), …
- The number under \(\Sigma\) is the lower limit (the first value of \(r\)).
- The number above \(\Sigma\) is the upper limit (the last value of \(r\)).
Read \(\displaystyle\sum_{r=1}^{4} 3r\) as “the sum of \(3r\) for \(r\) from 1 to 4”.
Example 1 Write out each series in full.
(a) \(\displaystyle\sum_{r=1}^{4} 3r\)
(b) \(\displaystyle\sum_{k=2}^{5} \left(k^2 + 1\right)\)
Put \(r = 1, 2, 3, \ldots\) (from the lower limit to the upper limit) into the general term.
(a) \(3(1) + 3(2) + 3(3) + 3(4) = 3 + 6 + 9 + 12\)
(b) \(k = 2, 3, 4, 5\): \(\;5 + 10 + 17 + 26\)
The number of terms in \(\displaystyle\sum_{r=m}^{n} u_r\) is \(n - m + 1\). For example, \(\displaystyle\sum_{r=5}^{24}\) has \(24 - 5 + 1 = 20\) terms.
Example 2 Find the value of each sum.
(a) \(\displaystyle\sum_{r=0}^{3} \frac{2}{r^2 + 1}\)
(b) \(\displaystyle\sum_{r=-1}^{2} (-1)^r\, 3^r\)
(a) \(r = 0, 1, 2, 3\):
\[ \frac{2}{1} + \frac{2}{2} + \frac{2}{5} + \frac{2}{10} = 2 + 1 + 0.4 + 0.2 = 3.6 \]
(b) \(r = -1, 0, 1, 2\):
\[ (-1)^{-1}3^{-1} + (-1)^0 3^0 + (-1)^1 3^1 + (-1)^2 3^2 = -\frac{1}{3} + 1 - 3 + 9 = \frac{20}{3} \]
Arithmetic and geometric series in sigma notation
If the general term is linear in \(r\) (like \(4r - 3\)), the series is arithmetic. If it is of the form \(a \times b^{r}\), the series is geometric. Then you can use the formulae from 2.2 and 2.3 instead of adding every term.
Example 3 Find \(\displaystyle\sum_{r=5}^{24} (4r - 3)\).
The series is arithmetic (\(d = 4\)). It has \(24 - 5 + 1 = 20\) terms.
First term (\(r = 5\)): \(17\). Last term (\(r = 24\)): \(93\).
\[ S_{20} = \frac{20}{2}(17 + 93) = 10 \times 110 = 1100 \]
Example 4 Find
(a) \(\displaystyle\sum_{r=1}^{10} 3 \times 2^{\,r-1}\)
(b) \(\displaystyle\sum_{r=1}^{\infty} 5\,(0.4)^{r}\)
(a) Geometric with \(u_1 = 3\), \(r = 2\), 10 terms: \(\;S_{10} = \dfrac{3\left(2^{10} - 1\right)}{2 - 1} = 3069\).
(b) Geometric with \(u_1 = 5 \times 0.4 = 2\) and ratio \(0.4\). Since \(|0.4| < 1\):
\[ \sum_{r=1}^{\infty} 5\,(0.4)^{r} = \frac{2}{1 - 0.4} = \frac{10}{3} . \]
Careful: the first term is the value at \(r = 1\), which is \(2\), not \(5\).
Rules of sigma notation
Rules of sigma notation Not in the formula booklet
| Rule | |
|---|---|
| Constant factor | \(\displaystyle\sum_{r=1}^{n} k\,u_r = k\sum_{r=1}^{n} u_r\) |
| Sum and difference | \(\displaystyle\sum_{r=1}^{n} (u_r \pm v_r) = \sum_{r=1}^{n} u_r \pm \sum_{r=1}^{n} v_r\) |
| Sum of a constant | \(\displaystyle\sum_{r=1}^{n} c = cn \qquad \left(\text{and } \sum_{r=m}^{n} c = c(n - m + 1)\right)\) |
| Changing the lower limit | \(\displaystyle\sum_{r=m}^{n} u_r = \sum_{r=1}^{n} u_r - \sum_{r=1}^{m-1} u_r\) |
- \(r\): the index (a counter)
- \(n\): the upper limit (the last value of \(r\))
- \(k\): a constant
- \(u_r\): the general term of a sum
- \(v_r\): the general term of another sum
- \(c\): a constant
- \(m\): the lower limit (the first value of \(r\))
These rules work because you can rearrange the terms of a finite sum. But products and quotients do not split:
\[ \sum u_r v_r \neq \left(\sum u_r\right)\left(\sum v_r\right), \qquad \sum \frac{u_r}{v_r} \neq \frac{\sum u_r}{\sum v_r} . \]
Example 5 The \(r\)th term of a series is \(2^{\,r-1} + 3r\). Find the sum of the first \(n\) terms.
Split the sum into a geometric part and an arithmetic part:
\[ \sum_{r=1}^{n} \left(2^{\,r-1} + 3r\right) = \sum_{r=1}^{n} 2^{\,r-1} + \sum_{r=1}^{n} 3r . \]
Geometric part (\(u_1 = 1\), ratio \(2\)): \(\;\dfrac{2^n - 1}{2 - 1} = 2^n - 1\).
Arithmetic part (\(3 + 6 + \cdots + 3n\)): \(\;\dfrac{n}{2}(3 + 3n) = \dfrac{3n(n + 1)}{2}\).
\[ \text{Total} = 2^n - 1 + \frac{3n(n + 1)}{2} \]
Writing a series in sigma notation
Find a general term that fits every term, and the limits that give the first and last terms.
Example 6 Write each series in sigma notation.
(a) \(3 + 7 + 11 + \cdots + 83\)
(b) \(\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} + \cdots + \dfrac{1}{3^8}\)
(c) \(2 - 4 + 8 - 16 + \cdots + 512\) \(\qquad\) (d) \(-1 + 4 - 9 + 16 - 25\)
(a) Arithmetic with \(d = 4\): general term \(4r - 1\). Last term: \(4r - 1 = 83 \Rightarrow r = 21\). \(\quad\displaystyle\sum_{r=1}^{21} (4r - 1)\)
(b) General term \(\left(\frac{1}{3}\right)^r\), from \(r = 1\) to \(8\): \(\quad\displaystyle\sum_{r=1}^{8} \left(\frac{1}{3}\right)^{r}\)
(c) The sizes are \(2^r\). The signs alternate, starting with \(+\), so use \((-1)^{r+1}\). \(512 = 2^9\), and \(r = 9\) gives a \(+\) sign ✓. \(\quad\displaystyle\sum_{r=1}^{9} (-1)^{r+1}\, 2^{r}\)
(d) The sizes are \(r^2\), and the first term is negative, so use \((-1)^r\). \(\quad\displaystyle\sum_{r=1}^{5} (-1)^{r}\, r^{2}\)
There is often more than one correct answer. For example, (a) can also be written \(\displaystyle\sum_{r=0}^{20} (4r + 3)\).
Sigma notation on the GDC
On a Calculator page, choose Menu → Calculus → Sum (or pick the \(\displaystyle\sum\) template from the templates palette). Use [tab] to move between the boxes: the index, the lower limit, the upper limit, and the general term. Press [enter].
For example, \(\displaystyle\sum_{r=7}^{20} (6r + 5)\) gives \(1204\).
Example 7 Use your GDC to find
(a) \(\displaystyle\sum_{r=1}^{25} (2 + r)(1 - r)\)
(b) \(\displaystyle\sum_{r=1}^{100} 0.99^{r}\)
(a) \(-5800\) \(\qquad\) (b) \(62.8\) (3 s.f.)
Check (b) by hand: it is geometric with \(u_1 = 0.99\) and ratio \(0.99\), so \(\dfrac{0.99\left(1 - 0.99^{100}\right)}{0.01} = 62.76\ldots\) ✓
Extension — sums of powers
These results are not in the syllabus or the formula booklet, but they are useful.
\[ \sum_{r=1}^{n} r = \frac{n(n + 1)}{2}, \qquad \sum_{r=1}^{n} r^2 = \frac{n(n + 1)(2n + 1)}{6}, \qquad \sum_{r=1}^{n} r^3 = \frac{n^2(n + 1)^2}{4} \]
The first one is an arithmetic series. The other two can be proved by mathematical induction (Unit 6).
Example. \(2^2 + 5^2 + 8^2 + \cdots + 29^2 = \displaystyle\sum_{r=1}^{10} (3r - 1)^2\). Expand and use the rules:
\[ \sum_{r=1}^{10} \left(9r^2 - 6r + 1\right) = 9 \times \frac{10 \times 11 \times 21}{6} - 6 \times \frac{10 \times 11}{2} + 10 = 3465 - 330 + 10 = 3145 . \]
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| \(\displaystyle\sum_{r=5}^{24}\) has 19 terms | 20 terms | The number of terms is \(n - m + 1\). |
| \(\displaystyle\sum_{r=1}^{\infty} 5(0.4)^r\) has first term 5 | first term \(2\) | Substitute the lower limit, \(r = 1\). |
| \(\displaystyle\sum_{r=1}^{n} 7 = 7\) | \(7n\) | You add the constant \(n\) times. |
| \(\displaystyle\sum r^2 = \left(\sum r\right)^2\) | not equal | Sigma does not split over products. |
| Signs \((-1)^r\) for \(2 - 4 + 8 - \cdots\) | \((-1)^{r+1}\) | Check the sign of the first term. |
Exercises
1 Write out in full and evaluate \(\displaystyle\sum_{r=1}^{5} (2r + 1)\).
\(3 + 5 + 7 + 9 + 11 = 35\).
2 Evaluate \(\displaystyle\sum_{k=0}^{4} \frac{(-1)^k}{2^k}\).
\(1 - \dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{8} + \dfrac{1}{16} = \dfrac{16 - 8 + 4 - 2 + 1}{16} = \dfrac{11}{16}\).
3 How many terms are in \(\displaystyle\sum_{r=12}^{50} u_r\)?
\(50 - 12 + 1 = 39\) terms.
4 Find \(\displaystyle\sum_{r=1}^{40} (5r - 2)\).
Arithmetic: first term \(3\), last term \(198\), 40 terms. \(\;\dfrac{40}{2}(3 + 198) = 4020\).
5 Find \(\displaystyle\sum_{r=10}^{30} (2r + 7)\).
21 terms; first \(27\), last \(67\). \(\;\dfrac{21}{2}(27 + 67) = 987\).
6 Find \(\displaystyle\sum_{r=1}^{8} 5 \times 3^{\,r-1}\).
\(\dfrac{5\left(3^8 - 1\right)}{3 - 1} = \dfrac{5 \times 6560}{2} = 16\,400\).
7 Find \(\displaystyle\sum_{r=1}^{\infty} 12\left(-\frac{1}{2}\right)^{r}\).
First term (\(r = 1\)): \(-6\); ratio \(-\dfrac{1}{2}\). \(\;\dfrac{-6}{1 + \frac{1}{2}} = -4\).
8 Write in sigma notation:
(a) \(5 + 8 + 11 + \cdots + 62\)
(b) \(1 - 3 + 9 - 27 + \cdots - 2187\)
(a) General term \(3r + 2\); \(3r + 2 = 62 \Rightarrow r = 20\): \(\;\displaystyle\sum_{r=1}^{20} (3r + 2)\).
(b) \((-3)^{r}\) from \(r = 0\): \((-3)^7 = -2187\), so \(\;\displaystyle\sum_{r=0}^{7} (-3)^{r}\).
9 Given that \(\displaystyle\sum_{r=1}^{n} (2r + 3) = 140\), find \(n\).
Arithmetic, first term \(5\), last term \(2n + 3\): \(\;\dfrac{n}{2}(5 + 2n + 3) = n(n + 4) = 140\).
\(n^2 + 4n - 140 = 0 \Rightarrow (n + 14)(n - 10) = 0\). Since \(n > 0\), \(n = 10\).
10 Find, in terms of \(n\), \(\displaystyle\sum_{r=1}^{n} \left(4r + 5^{r}\right)\).
\(\displaystyle\sum_{r=1}^{n} 4r = \frac{n}{2}(4 + 4n) = 2n(n + 1)\) and \(\displaystyle\sum_{r=1}^{n} 5^{r} = \frac{5\left(5^n - 1\right)}{4}\).
Total: \(2n(n + 1) + \dfrac{5\left(5^{n} - 1\right)}{4}\).