26.1 Sequences

A sequence is a lot of numbers, one after the other. Usually the terms are related in some way to the previous terms. For example, in this sequence11 1 Known as the Fibonacci sequence, each term is the sum of the two previous terms22 2 If we set that the first term is 11 and the second 22 then to get the third term, we add the previous two, so 1+2=31+2=3. This process continues for the rest of the terms in the sequence.

1,2,3,5,8,13,1,2,3,5,8,13,... (26.1)

We can also write the nth term of a sequence xx as x(n)x(n), and denote the entire sequence as {xn}\{x_{n}\}

26.1.1 Convergence of sequences

Sometimes a sequence can converge, which informally means that the value of the nnth term in the sequence gets closer and closer to a specific value - infinitely close in fact!

Can we say more than \saythis sequence looks like it’s getting closer and closer to this value though? Yes! Another way of saying that something is getting \saycloser to a value, is to say that the distance between the two is decreasing. Using the modulus function 33 3 There is a section on the modulus function in the ”Algebra” chapter of this document, and another section on how we can write distances in terms of the modulus function in the vectors chapter. we can write the distance between any two points x(n)x(n) and aa as

|a(n)a|\left\lvert a(n)-a\right\rvert (26.2)

What about "really close"? The way to do this is to say that for every value of ϵ>0\epsilon>0 where ϵ\epsilon is a real number and as nn\to\infty we have that

|x(n)a|<ϵ\left\lvert x(n)-a\right\rvert<\epsilon (26.3)

then {xn}\{x_{n}\} converges to aa. As we can pick any ϵ>0\epsilon>0 (but not ϵ=0\epsilon=0) what we are in effect saying is that the distance is infinitely close to zero, but not equal to zero. That is, it converges to 0 but is not equal to zero.

26.1.2 A sequence can only converge to one value

Before proving this, it’s helpful to formulate precisely what we want to prove.

What 44 4 Note that there exists is the negation of for every. It helps me to imagine a haystack. If you tell me that \saythere exists a needle in this haystack, then the only way I can disprove you is to search through all of (for every) the haystack and show that there is no needle. If I say \sayfor every piece of hay in this haystack, none of them is hiding a needle, then you can disprove me by showing that at least one piece of hay is hiding a needle.we want to show is that for every converging sequence, there exists a unique value to which it converges. The only way this could not be true is if there exists at least one sequence which converges to more than one value.

To prove this by contradiction, let us assume that there is indeed a sequence {xn}\{x_{n}\} which converges to both aa and bb, where aba\neq b. In this case we have that for all ϵ>0\epsilon>0 that as nn\to\infty,

|xna|<ϵ and |xnb|<ϵ\displaystyle\left\lvert x_{n}-a\right\rvert<\epsilon\text{ and }\left\lvert x% _{n}-b\right\rvert<\epsilon (26.4)

We would like to show that a=ba=b, or that ab=0a-b=0 (as this would be a contradiction). We know that

|ab|=|axn+xnb|\left\lvert a-b\right\rvert=\left\lvert a-x_{n}+x_{n}-b\right\rvert (26.5)

Using the triangle inequality we can write that

55 5 Unfortunately, this equation is a bit ambiguous - what I’m trying to say is that everything on the left-hand side is less than or equal to everything on the right-hand side and that everything on the right-hand side is equal.
|axn+xnb|\displaystyle\left\lvert a-x_{n}+x_{n}-b\right\rvert |axn|+|xnb|\displaystyle\leqq\left\lvert a-x_{n}\right\rvert+\left\lvert x_{n}-b\right\rvert (26.6)
=ϵ+ϵ\displaystyle=\epsilon+\epsilon (26.7)
=2ϵ\displaystyle=2\epsilon (26.8)

Therefore, overall we have that

|ab|2ϵ\left\lvert a-b\right\rvert\leqq 2\epsilon (26.9)

However, we can pick any value which is greater than 0 for ϵ\epsilon here. If as we are assuming, aba\neq b then for any constant c>0c>0 we also have that c|ab|c\left\lvert a-b\right\rvert is greater than 0. If we set cc to something less than 12\frac{1}{2} (e.g. 13\frac{1}{3}) then we would have that

|ab|23|ab|\left\lvert a-b\right\rvert\leqq\frac{2}{3}\left\lvert a-b\right\rvert (26.10)

This is definitely not true, and hence we have derived a contradiction; thus there can only be one.