Sigma Notation and Partial Sums, precalculus lesson

Adding the first twenty squares means writing twenty numbers and nineteen plus signs. Sigma notation packs that whole sum into one short expression that says what to add, where to start and where to stop. It also gives a clean way to keep running totals, which is how infinite sums will later get a meaning.

Key terms

Finite sum in sigma notation

Definition Let $m \le n$ be integers and let $a_m, a_{m+1}, \dots, a_n$ be real numbers. Their finite sum is written

\sum_{k=m}^{n} a_k = a_m + a_{m+1} + \cdots + a_n.

The symbol $\Sigma$ is the capital Greek letter sigma. A finite sum is a single real number, and since addition can be reordered and regrouped freely, its value does not depend on the order in which the terms are added.

In plain words Put each whole number from the bottom value to the top value into the formula and add the results.

Example The first five squares add up to

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

Summation index

Definition The letter under the sigma, $k$ in the sum above, is the summation index. It takes every integer value from $m$ to $n$ , each exactly once, in steps of $1$ . Common choices are $i$ , $j$ , $k$ and $n$ .

In plain words The index is a counter that always moves up by one. If you want to skip numbers, the skipping has to be built into the formula.

Example To add the even numbers from $2$ to $10$ the index cannot jump by $2$ . Instead the even numbers are written as $2k$ ,

\sum_{k=1}^{5} 2k = 2 + 4 + 6 + 8 + 10 = 30.

Summand

Definition The expression $a_k$ that follows the sigma is the summand, also called the general term. It is the rule that gives a term for each value of the index. A summand that does not involve the index is constant, and then every term is the same.

In plain words The summand is the formula for a typical term. Put it in parentheses when it has more than one piece, so it is clear what is being added.

Example Parentheses decide what the sigma covers.

\begin{gathered} \sum_{k=1}^{4} (2k + 1) = 3 + 5 + 7 + 9 = 24, \\[0.8em] \left(\sum_{k=1}^{4} 2k\right) + 1 = 20 + 1 = 21. \end{gathered}

Lower and upper bounds

Definition In the sum of $a_k$ from $k = m$ to $k = n$ , the integer $m$ is the lower bound and $n$ is the upper bound, also called the lower and upper limits of summation. Both bounds are included, so the sum has $n - m + 1$ terms. For a constant $c$ this gives

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

In plain words Count the terms as top minus bottom plus one. Forgetting the plus one is the most common slip.

Example This sum has $9 - 4 + 1 = 6$ terms,

\begin{aligned} \sum_{k=4}^{9} (k - 1) &= 3 + 4 + 5 + 6 + 7 + 8 \\[0.4em] &= 33. \end{aligned}

With the constant summand $5$ and the same bounds the sum is $6 \cdot 5 = 30$ .

Dummy index

Definition The summation index is a dummy index, also called a bound variable. The value of the sum does not depend on its name,

\sum_{k=m}^{n} a_k = \sum_{j=m}^{n} a_j,

and the index has no meaning outside the sum. Other letters in a sum, such as $n$ in an upper bound or a constant $x$ in the summand, are not bound, and the value of the sum can depend on them.

In plain words You can rename the counter however you like. The answer may depend on the bounds and on other letters, but never on the counter.

Example Both sums below equal $1 + 8 + 27 = 36$ ,

\sum_{k=1}^{3} k^3 = \sum_{j=1}^{3} j^3 = 36.

In the next sum the answer depends on $x$ and not on $k$ , so a $k$ in the result would make no sense,

\begin{aligned} &\sum_{k=1}^{3} (k + x) \\[0.4em] &\quad = (1 + x) + (2 + x) + (3 + x) \\[0.4em] &\quad = 6 + 3x. \end{aligned}

Sum rules

Definition For real numbers $a_k$ and $b_k$ and a constant $c$ ,

\begin{gathered} \sum_{k=m}^{n} c\,a_k = c \sum_{k=m}^{n} a_k, \\[0.8em] \sum_{k=m}^{n} (a_k \pm b_k) = \sum_{k=m}^{n} a_k \pm \sum_{k=m}^{n} b_k. \end{gathered}

For an integer $p$ with $m \le p < n$ a sum also splits into two consecutive pieces,

\sum_{k=m}^{n} a_k = \sum_{k=m}^{p} a_k + \sum_{k=p+1}^{n} a_k.

There are no such rules for products, quotients or powers. In general the sum of the products $a_k b_k$ is not the product of the two sums.

In plain words Constants factor out, sums and differences split term by term, and a long sum can be cut into two shorter ones. All of this is the same additions regrouped.

Example Since the numbers $1$ to $4$ add up to $10$ and their squares add up to $30$ ,

\begin{aligned} \sum_{k=1}^{4} (3k^2 - 2k) &= 3 \cdot 30 - 2 \cdot 10 \\[0.4em] &= 70. \end{aligned}

Adding directly gives $1 + 8 + 21 + 40 = 70$ as well. Splitting helps with sums that start late. The numbers $1$ to $10$ add up to $55$ , so the numbers $5$ to $10$ add up to $55 - 10 = 45$ . For products the rule fails already with two terms, since $1 \cdot 1 + 2 \cdot 2 = 5$ while $(1 + 2)(1 + 2) = 9$ .

Reindexing

Definition For an integer $p$ , the substitution $k = j + p$ gives

\sum_{k=m}^{n} a_k = \sum_{j=m-p}^{n-p} a_{j+p}.

The bounds move by $p$ in one direction and the summand moves to compensate, so exactly the same terms are added.

In plain words You can slide the counter to start wherever is convenient, as long as you change the formula to match. Write out the first and last terms of both versions to check.

Example The substitution $k = j + 3$ moves the start from $3$ to $0$ ,

\begin{aligned} \sum_{k=3}^{7} 2^{k-3} &= \sum_{j=0}^{4} 2^{j} \\[0.8em] &= 1 + 2 + 4 + 8 + 16 \\[0.4em] &= 31. \end{aligned}

Both versions have five terms, starting with $1$ and ending with $16$ .

Partial sum

Definition Let $\{a_k\}$ be a sequence with first index $1$ . Its $n$ th partial sum is

S_n = \sum_{k=1}^{n} a_k = a_1 + a_2 + \cdots + a_n.

The partial sums form a new sequence $\{S_n\}$ . They satisfy $S_1 = a_1$ and $S_n = S_{n-1} + a_n$ for $n \ge 2$ , so the terms can be recovered from the partial sums as $a_1 = S_1$ and $a_n = S_n - S_{n-1}$ for $n \ge 2$ . When the first index is $0$ the running total starts at $a_0$ , and $S_n$ is still the sum of the first $n$ terms, $a_0 + a_1 + \cdots + a_{n-1}$ , as in the calculus lessons. Some books use $a_0 + \cdots + a_n$ instead, so check which one is meant.

In plain words Keep a running total. The partial sum $S_n$ is the total after the first $n$ terms.

Example For the odd numbers $a_k = 2k - 1$ the running totals look like this.

$n$	$1$	$2$	$3$	$4$	$5$
$a_n$	$1$	$3$	$5$	$7$	$9$
$S_n$	$1$	$4$	$9$	$16$	$25$

Every partial sum is a perfect square, which suggests $S_n = n^2$ . A pattern like this can be proved for every $n$ by mathematical induction.

Sequence versus series

Definition A sequence is an ordered list of numbers $a_1, a_2, a_3, \dots$ . A series is the sum of the terms of a sequence, written $a_1 + a_2 + a_3 + \cdots$ or in sigma notation. A finite series is an ordinary finite sum. An infinite series

\sum_{k=1}^{\infty} a_k = a_1 + a_2 + a_3 + \cdots

has infinitely many terms, so its value cannot come from adding them one by one. If the partial sums $S_n$ approach a number $S$ as $n$ grows, the series converges and its sum is $S$ . Otherwise the series diverges.

In plain words A sequence is the list. A series adds the list up, and an infinite series gets its value from its running totals.

Example The sequence $1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \dots$ has terms that shrink toward $0$ . The series built from it has partial sums $1, 1.5, 1.75, 1.875, 1.9375$ , and in general

\begin{aligned} S_n &= 1 + \frac{1}{2} + \cdots + \frac{1}{2^{n-1}} \\[0.8em] &= 2 - \frac{1}{2^{n-1}}, \end{aligned}

which approaches $2$ . So the terms approach $0$ while the series has sum $2$ , two different questions with two different answers. In the same way the constant sequence $1, 1, 1, \dots$ converges to $1$ , but the series $1 + 1 + 1 + \cdots$ has partial sums $S_n = n$ and diverges. The lesson on infinite geometric series returns to this.

Check yourself

Write $3 + 7 + 11 + 15 + 19$ in sigma notation and find its value.

Show the answer

Each term is $4$ more than the one before and the first is $3$ , so the terms are $4k - 1$ for $k = 1$ to $5$ ,

\sum_{k=1}^{5} (4k - 1) = 55.

Rewrite the sum below so that its index $j$ starts at $1$ , then evaluate it.

\sum_{k=3}^{6} (k - 2)^3

Show the answer

Put $k = j + 2$ . The bounds become $1$ and $4$ and the summand becomes $j^3$ ,

\sum_{j=1}^{4} j^3 = 1 + 8 + 27 + 64 = 100.

The partial sums of a sequence are $S_n = n^2 + 3n$ for $n \ge 1$ . Find $a_1$ , $a_5$ and a formula for $a_n$ .

Show the answer

First $a_1 = S_1 = 4$ and $a_5 = S_5 - S_4 = 40 - 28 = 12$ . For $n \ge 2$

\begin{aligned} a_n &= S_n - S_{n-1} \\[0.4em] &= \left(n^2 + 3n\right) \\ &\quad - \left((n - 1)^2 + 3(n - 1)\right) \\[0.4em] &= 2n + 2, \end{aligned}

and this also gives $a_1 = 4$ , so $a_n = 2n + 2$ for every $n \ge 1$ .

Sigma Notation and Partial Sums

Key terms

Finite sum in sigma notation

Summation index

Summand

Lower and upper bounds

Dummy index

Sum rules

Reindexing

Partial sum

Sequence versus series

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus