Wednesday, 9 November 2011

Sequences and Series

In maths a sequence is an ordered list of numbers called objects, a sequence can be finite, infinite or bi-infinite. A finite sequence is easy to define it is simply a set of finite numbers (for example 1, 3, 5, 7 is a finite sequence); an infinite sequence is slightly harder to picture but it also makes sense, it is simply a sequence that does not terminate at any point (for example the natural numbers start are 1, 2, 3... and will never cease) and a bi-infinite sequence may be a slightly newer concept to try and grasp but it is equally easy to understand, it stretches infinitely in both directions (for example the integers are a bi-infinite sequence ...-2, -1, 0, 1, 2... as it will never cease in either direction).

The sum of a sequence is a series, and this is where the real fun of it all begins to come through with sequences. Often we look to see whether a series is divergent, convergent or oscillating, what this means is whether the series will converge to a limit, if it is divergent it will just get larger and larger in either the positive or negative direction and if a series is oscillating it neither converges to a limit nor diverges to infinity (for example: 1, -1, 1, -1, ... will not converge or diverge, thus it is oscillating).

It is also possible to find the sum of an arithmetic and geometric series (these will be defined later), but first I'll have to introduce some concepts of the symbols used when talking of the sums of numbers. If we have a function, f(x), our first term is, r, and we want to go up to term, n. The way you would usually do this long hand and have to do f(r)+f(r+1)+...+f(n-1)+f(n), but this can be more concisely wrote as:

As you can see this is far easier to write and is just as easy to interpret. But how actually do you find the sum of a series?

For example if we wanted to find the sum to the first 10 natural numbers, 1+2+3+4+5+6+7+8+9+10, what way could we find the answer relatively quickly? Pairing the numbers up first and last, second and second last, etc. will mean each number will be the same so makes an easier calculation. If we do that we get (1+10)+(2+9)+(3+8)+(4+7)+(5+6), which is 11+11+11+11+11 (or better represented as 11*5). So the sum of the first 10 natural numbers is 55.

We can write this short hand by finding out how many pairs their will be and what each one will equal, or S = 0.5n(a+l), where n is the number of terms, a is the first term and l is the last term. Another way of writing this (when we do not know the last term) is that l = a+(n-1)d, if you want an explanation on this please comment. This then means that the sum of an arithmetic series is, S = 0.5n(2a+(n-1)d). **Note that these last two paragraphs only apply to arithmetic series**

We can also find the sum of a geometric sequence (a term starts with 'a' and is the increased by 'r', so the next will be 'ar' and the third will be 'ar²' etc.). This means that the nth term of the sequence will be, arn-1. So to find the sum of a geometric progression, let us consider that Sn = a+ar+ar²+...+arn-1, if we multiply through by r we get rSn = ar+ar²+...+arn, if we take the first from the second we get: Sn-rS= a-arn, which can be represented as Sn(1-r= a-arn, dividing through by (1-r) we get the formula for the sum of a geometric progression:

Now we can use this to find the sum to infinity of geometric progression (provided that -1<r<1), it will mean that will arn approach 0 as n approaches infinity, when we have that it becomes easy to find the new equation:

There are also other elements to this general topic, but I think they warrant a topic in their own right. I briefly cover the definition of what a limit is here.

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