26.1 Sequences

Sometimes a sequence can converge, which informally means that the value of the $n$th 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 $x(n)$ and $a$ as

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

then $\{x_{n}\}$ converges to $a$. As we can pick any $\epsilon>0$ (but not $\epsilon=0$) 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.

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 $\{x_{n}\}$ which converges to both $a$ and $b$, where $a\neq b$. In this case we have that for all $\epsilon>0$ that as $n\to\infty$,

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

Using the triangle inequality we can write that

Therefore, overall we have that

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

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