In this section, I'll review the notation for sums and products.

Addition and multiplication are * binary
operations*: They operate on two numbers at a time. If you want
to add or multiply more than two numbers, you need to group the
numbers so that you're only adding or multiplying two at once.

Addition and multiplication are associative, so it should not matter how I group the numbers when I add or multiply. Now associativity holds for three numbers at a time; for instance, for addition,

But why does this work for (say) two ways of grouping a sum of 100 numbers?

It turns out that, using * induction*, it's
possible to prove that any two ways of grouping a sum or product of n
numbers, for , give the same result.

Taking this for granted, I can define the sum of a list of numbers
*
recursively* as follows.

A sum of a single number is just the number:

I know how to add two numbers:

To add n numbers for , I add the first of them (which I assume I already know how to do), then add the :

Note that this is a particular way of grouping the n summands: Make one group with the first , and add it to the last term. By associativity (as noted above), this way of grouping the sum should give the same result as any other way.

Therefore, I might as well omit parentheses and write

The product of a list of numbers
is defined * recursively* in a similar way.

A product of a single number is just the number:

I know how to multiply two numbers:

To multiply n numbers for , I multiply the first of them (which I assume I already know how to do), then multiply the :

Again, I can drop the parentheses and write

* Remark.* Some people define the sum and product
of *no* numbers as follows: The sum of no numbers (sometimes
called the * empty sum* is defined to be 0, and
the product of no numbers (sometimes called the *
empty product*) is defined to be 1. These definitions are useful
when you're taking the sum or product of numbers in a set, rather
than the sum or product of numbers indexed by a range of consecutive
integers.

For example, suppose . Then to use summation notation to write the sum of the squares of the elements of S, you might write

Similarly, to write the product of the reciprocals of the numbers in S, you might write

If the set containing the elements of the sum or product varies in a discussion, it might happen that the containing set is empty. Using the definitions for an empty sum or an empty product allows for this case. For instance,

That is, the sum of the squares of the elements of the empty set is 0.

The following properties and those for products which are given below are fairly obvious, but careful proofs require induction.

* Proposition.* (* Properties of
sums*)

(a) .

(b) .

(c) .

* Proof.* I'll prove (c) as an example. I'll use
induction. For , the left side is and the right side is . The result holds for .

Assume that the result holds for n:

Then for , I have

The first equality used the recursive definition of a sum. The second equality follows from the induction hypothesis. The final equality is just basic algebra.

Properties (a) and (b) can also be proved using induction.

* Example.* Compute the following sums:

(a) .

(b) .

(c) .

(d) , where c is a constant.

(a)

(b)

Note that sums need not start indexing at 1.

(c)

Note that the summation variable can be anything that you want.

(d)

* Example.* Compute the exact value of .

Note that

So

I'll write out a few of the terms:

Most of the terms cancel in pairs, with the negative term in one expression cancelling with the positive term in the next expression. Only the first term and the last term are left, and they give the value of the sum:

This is an example of a * telescoping sum*.

* Proposition.* (* Properties of
products*)

(a) .

(b) .

(c) .

* Proof.* I'll prove (a) by way of example. The
proof will use induction on n.

For , the left side is . The right side is . The result holds for .

Assume that the result holds for n:

Then for , I have

The first equality follows from the recursive definition of a product. The second equality uses the induction hypothesis. The third equality comes from commutativity of multiplication. And the fourth equality is a result of the recursive definition of a product again.

Properties (b) and (c) can also be proved by induction.

* Example.* Compute the following products:

(a) .

(b) , where c is a constant.

(c) .

(a)

(b)

(c)

* Example.* Compute the exact value of .

Note that

So

I'll write out a few of the terms:

Most of the terms in the numerators cancel with the terms in the denominators "two fractions ahead". Only the first two denominators and the last two numerators don't cancel, and they give the value of the product:

This is an example of a * telescoping
product*.

If n is a positive integer, (read *
n-factorial*) is the product of the integers from 1 to n:

For example,

By convention, is defined to be 1. This definition agrees
with the extension of factorials using the Gamma function, described
below. It also is the appropriate definition when factorials come up
in * binomial coefficients*.

Note that

There is a way of extending the definition of factorials to positive
real numbers (so that, for instance, is defined). The * Gamma
function* is defined by

(Note that x here is a *real number*.) If n is a positive
integer,

Here's a graph of the Gamma function:

I've placed dots at the points so you can see that the Gamma function really goes through the factorial points. Note also that . But if I plug into the formula above, I get , i.e. . Thus, , which is the convention I mentioned earlier.

* Example.* Evaluate using the integral definition.

Do the antiderrivative by parts:

So the definite integral is

All the b-terms go to 0, because

Note that , so .

Copyright 2019 by Bruce Ikenaga