## 30 Nov 2008

### presentation of Edwards & Penney's work, by by Corry Shores[Search Blog Here. Index-tags are found on the bottom of the left column.][Central Entry Directory][Mathematics, Calculus, Geometry, Entry Directory][Calculus Entry Directory][Edwards & Penney, Entry Directory]

Edwards & Penney's Calculus is an incredibly-impressive, comprehensive, and understandable book. I highly recommend it.

An infinite series is an infinite sequence whose terms are summed, and it takes the form:

here, {an} is an infinite sequence of real numbers, and the number an is the nth term of the series. We may abbreviate the left side of the equation with:

We will now consider the sums of infinite sequences, for example:

We cannot easily sum an infinity of terms, but we may add a finite sequence, so the sum of the first five terms of this sequence is:

We could continue adding five more terms at a time, and we see how this progresses:

We see that the series appears to get closer and closer to 1 as we continually add more terms. Thus we might say that the sum of the whole infinite series is 1:

But we may at least determine sums of a finite part of an infinite sequence, that is, to find partial sums of it.

Thus we may say that infinite series are made up not merely of infinite sequences of terms, but as well of infinite sequences of partial sums:

and so on. The sum of an infinite series is defined as the limit of its sequence of partial sums, so long as it actually has such a limit.

exists (and is finite). Otherwise we say that the series diverges (or is divergent). If a series diverges, then it has no sum.

Thus, so long as it has a limit, the sum of an infinite series is a limit of finite sums:

For example, we might want to find the sum of -- and show that there is a convergence in -- the series:

The first four partial sums of this series are:

We see that in each case, the numerator is n-squared minus 1, over n-squared, so it seem probable that

In fact, we know that this is so by induction:

For this reason, the sum of our series is:

We arrive at the middle equation by setting 2^n over 2^n, and negative 1 over 2^n. Then as 1/2^n goes to infinity, it becomes 0, hence to subtract it from 1 leaves 1. We see below a graph of the partial sums of this series:

from Edwards & Penney: Calculus. New Jersey: Prentice Hall, 2002, p692-693.