Sums and Sigma Notation, calculus lesson

Adding the first hundred whole numbers by hand takes a while, and writing the sum out takes nearly as long. Areas under curves will soon be computed as sums of many thin rectangles, with the number of rectangles growing without bound. A short notation for such sums, plus a few formulas that add them up exactly, is what makes that possible.

Key terms

Sigma notation

Definition Let $m \le n$ be integers and let $a_m, a_{m+1}, \ldots, a_n$ be real numbers. Then

\sum_{i=m}^{n} a_i = a_m + a_{m+1} + \cdots + a_n.

The symbol $\Sigma$ is the capital Greek letter sigma, standing for sum.

In plain words The notation says to plug in each whole number from the bottom value to the top value and add up the results.

Example

\sum_{i=1}^{4} i^2 = 1 + 4 + 9 + 16 = 30.

Index of summation

Definition The variable $i$ in the sum above is the index of summation. It runs through the integers $m, m + 1, \ldots, n$ . It is a dummy variable, meaning the value of the sum does not depend on which letter is used,

\sum_{i=m}^{n} a_i = \sum_{k=m}^{n} a_k.

In plain words The index is just a counter. Renaming it changes nothing, and it has no meaning outside the sum.

Example Both of these sums equal $30$ ,

\sum_{i=1}^{4} i^2 = \sum_{k=1}^{4} k^2 = 30.

But in the sum below the summand does not involve the index at all, so it simply adds four copies of $k^2$ ,

\sum_{i=1}^{4} k^2 = 4k^2.

Summand

Definition In the sum of $a_i$ from $i = m$ to $n$ , the expression $a_i$ is the summand, or general term. It is a rule that gives a number for each value of the index.

In plain words The summand is the formula for a typical term.

Example With summand $2k + 1$ ,

\sum_{k=0}^{3} (2k + 1) = 1 + 3 + 5 + 7 = 16.

Limits of summation

Definition In the sum of $a_i$ from $i = m$ to $n$ , the integer $m$ is the lower limit and $n$ is the upper limit of summation. The sum has $n - m + 1$ terms. In particular, for a constant $c$ ,

\sum_{i=m}^{n} c = (n - m + 1)c.

In plain words The limits say where the counting starts and stops. Count the terms as top minus bottom plus one.

Example This sum has $7 - 3 + 1 = 5$ terms,

\sum_{j=3}^{7} j = 3 + 4 + 5 + 6 + 7 = 25.

A constant summand $4$ with $i$ running from $1$ to $10$ gives $10 \cdot 4 = 40$ .

Reindexing

Definition For any integer $p$ , substituting $i = j + p$ gives

\sum_{i=m}^{n} a_i = \sum_{j=m-p}^{n-p} a_{j+p}.

Both the limits and the summand change, and the list of terms being added stays the same.

In plain words You can slide the counter so that it starts where you like, as long as you adjust the formula to match.

Example With $i = j + 2$ ,

\sum_{i=3}^{6} (i - 2)^2 = \sum_{j=1}^{4} j^2 = 30.

Both sides are $1 + 4 + 9 + 16$ .

Sum rules

Definition For a constant $c$ and real numbers $a_i$ , $b_i$ ,

\sum_{i=m}^{n} c\,a_i = c\sum_{i=m}^{n} a_i, \qquad \sum_{i=m}^{n} (a_i \pm b_i) = \sum_{i=m}^{n} a_i \pm \sum_{i=m}^{n} b_i.

There is no corresponding rule for products. In general

\sum_{i=m}^{n} a_i b_i \neq \left(\sum_{i=m}^{n} a_i\right)\left(\sum_{i=m}^{n} b_i\right).

In plain words Constants factor out and sums split term by term, because addition can be done in any order.

Example The whole numbers from $1$ to $10$ add up to $55$ , so

\sum_{i=1}^{10} (3i + 2) = 3\sum_{i=1}^{10} i + \sum_{i=1}^{10} 2 = 3 \cdot 55 + 20 = 185.

For products the rule fails already with two terms,

\sum_{i=1}^{2} i \cdot i = 1 + 4 = 5, \qquad \left(\sum_{i=1}^{2} i\right)^2 = 3^2 = 9.

Sums of powers

Definition For every positive integer $n$ ,

\sum_{i=1}^{n} i = \frac{n(n + 1)}{2}, \qquad \sum_{i=1}^{n} i^2 = \frac{n(n + 1)(2n + 1)}{6}, \qquad \sum_{i=1}^{n} i^3 = \left(\frac{n(n + 1)}{2}\right)^2.

Each formula can be proved by induction on $n$ .

In plain words These closed forms add up the first $n$ whole numbers, squares or cubes in one step, however large $n$ is.

Example The first ten squares add up to

\sum_{i=1}^{10} i^2 = \frac{10 \cdot 11 \cdot 21}{6} = 385.

The formula also handles a sum that looks ahead to areas,

\frac{1}{n^3}\sum_{i=1}^{n} i^2 = \frac{(n + 1)(2n + 1)}{6n^2} \to \frac{2}{6} = \frac{1}{3} \quad \text{as } n \to \infty.

Finite geometric sums

Definition For a real number $r \neq 1$ , a real number $a$ and a positive integer $n$ ,

\sum_{i=0}^{n-1} a r^i = a + ar + \cdots + ar^{n-1} = a\,\frac{1 - r^n}{1 - r}.

If $r = 1$ the sum is $na$ . To prove the formula, call the sum $S$ . Then $S - rS = a - ar^n$ , because all the other terms cancel, and dividing by $1 - r$ gives the result.

In plain words When each term is a fixed multiple of the one before, the whole sum has a short formula. Remember it as the first term times $1 - r^n$ over $1 - r$ , where $n$ is the number of terms.

Example With $a = 3$ , $r = 2$ and $n = 6$ ,

\sum_{i=0}^{5} 3 \cdot 2^i = 3 \cdot \frac{1 - 2^6}{1 - 2} = 3 \cdot 63 = 189,

which matches $3 + 6 + 12 + 24 + 48 + 96 = 189$ .

Check yourself

Write $2 + 4 + 6 + \cdots + 20$ in sigma notation and find its value.

Show the answer

The terms are $2i$ for $i = 1$ to $10$ , so

\sum_{i=1}^{10} 2i = 2 \cdot \frac{10 \cdot 11}{2} = 110.

Find $\sum_{i=1}^{20} (i^2 - i)$ .

Show the answer

Split the sum and use the power formulas,

\sum_{i=1}^{20} i^2 = \frac{20 \cdot 21 \cdot 41}{6} = 2870, \qquad \sum_{i=1}^{20} i = \frac{20 \cdot 21}{2} = 210,

so the sum is $2870 - 210 = 2660$ .

Find $\sum_{k=1}^{6} \left(\frac{1}{2}\right)^k$ .

Show the answer

It is a geometric sum with first term $1/2$ , ratio $1/2$ and $6$ terms, so it equals

\frac{1}{2} \cdot \frac{1 - (1/2)^6}{1 - 1/2} = 1 - \frac{1}{64} = \frac{63}{64}.

Sums and Sigma Notation

Theory, part 93

Practice, part 94

Key terms

Sigma notation

Index of summation

Summand

Limits of summation

Reindexing

Sum rules

Sums of powers

Finite geometric sums

Check yourself

How this lesson connects

Used later in