Math 21, Fall, 2000

Lehigh University

5.0  Sigma notation

We will be spending quite a bit of time dealing with integration as a summation process, and it will get awkward writing down all these sums. We need a better notation. Usually the sums we will be ``casting up'' are indexed, that is, there is a whole number for each term in the sum, beginning with either 0 or 1 (sometimes higher), and ending at some larger value, or at infinity. So, instead of writing

a1+a2+a3+... +an,

we will write

n å i = 1  ai,

meaning the same thing. We call the second line sigma notation. Sometimes we use k or j or some other letter instead of i , but it doesn't matter, since it's just a ``dummy index'' of summation, like the name of a variable in the expression for a function - it's not the name of the variable, but the formula, that matters. The i or whatever often is involved in the expression for a_i , like a function. For example; let's look at:

4 å j = 1  1 j = 1 1 + 1 2 + 1 3 + 1 4 = 25 12 .

The sum

4 å k = 1  1 k

is exactly the same thing, 1/1+1/2+1/3+1/4 = 25/12 . Other examples:

Exercise 1 Compute

3 å i = 1  2i =

Exercise 2 Find

10 å n = 1  1 =

Think about this for a bit.

Exercise 3 Find

6 å k = 2  (-1)k =

Example 1 Find

100 å i = 2  ( 1 i - 1 i-1 ).

Solution

Summation formulas

The way I heard the story, little Carl Gauss was a bit of a discipline problem in school. As punishment one day, his teacher sent him to add the first thousand integers together. Little Carl came back rather sooner than the teacher had expected, with an answer. The teacher thought it was impossible to add that many numbers together that quickly. But, of course, little Carl was no ordinary discipline problem. He had figured out the following general formula:

n å i = 1  i = (n)(n+1) 2 ,

or, without the sigma:

1+2+3+4+5+... +n = (n)(n+1) 2 ,

and, since Gauss grew up to be one of the greatest mathematicians of all time, we can suppose that he showed his teacher the right sum, 500,500.

How did he figure that out? I suppose he used some sort of mathematical induction. He reasoned this way: he could easily show that the formula was true for n = 1 . He could also show that, if the formula is true for some integer n , then it is true for the next integer, which is n+1 . Thus; it's true for 1. But, if it's true for 1 it's true for 2. If it's true for 2 it's true for 3, etc. Where could it be false? It can't be false for any n . That is the essence of an induction proof; you need a place to start, where the statement is true, and you need to be able to show the next step, if you assume that the previous level is true.

Let's try it here:

1 å i = 1  i = 1,

which is the same as

1 = (1)(1+1) 2 ,

so the formula is true for n = 1 . Now, assume that, for a given n , that the formula is true:

n å i = 1  i = (n)(n+1) 2 .

We have to show that it's true for the next value of n :

n+1 å i = 1  i = (n+1)((n+1)+1) 2 .

The way you do that is to look at the sum:

n+1 å i = 1  i = æ è n å i = 1  i ö ø +(n+1) = n(n+1) 2 +(n+1) = n(n+1) 2 + 2(n+1) 2 = n(n+1)+2(n+1) 2 = (n+2)(n+1) 2 = (n+1)((n+1)+1) 2 ,

which is the same formula, but for n+1 . As little Gauss did, put it together and it says that that formula is true for all n .

There are two other formulas like this that come in handy:

n å i = 1  i2 = n(n+1)(2n+1) 6 ,

and

n å i = 1  i3 = é ê ë n(n+1) 2 ù ú û 2   ,

which can be proved the same way. There is a formula like this for the sum of the k^th powers of the first n integers, for any k , but these are the only ones commonly used. The one pattern that continues for all k is that the formula always involves the (k+1)^st power of n . The curious coincidence between the first powers and the third powers is not, so far as I know, repeated.

Of course, with modern computers we can all add more rapidly than Gauss' teacher. But these general formulas still play a significant role, as we'll see.

Example 2 Find

n å i = 1  3 n æ ç è æ ç è i n ö ÷ ø 2   +1 ö ÷ ø .

Solution

Exercise 4 Find

n å j = 1  ( j2+3j+1) =

Example 3 Evaluate

m å i = 1  æ è n å j = 1  ( i+j) ö ø .

Solution

Exercise 5 Find

lim n® ¥  æ ç è n å i = 1  1 n æ ç è i n ö ÷ ø 2   ö ÷ ø =

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