# 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 sequence^{1}^{1}
1
Known as the Fibonacci sequence, each term is the sum of the two
previous terms^{2}^{2}
2
If we set that the first term is $1$ and the second $2$
then to get the third term, we add the previous two, so $1+2=3$. This
process continues for the rest of the terms in the sequence.

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

## 26.1.1 Convergence of 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
^{3}^{3}
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.

## 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 ^{4}^{4}
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 $n\to\infty$,

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

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

Using the triangle inequality we can write that

^{5}

^{5}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.

$\displaystyle\left\lvert a-x_{n}+x_{n}-b\right\rvert$ | $\displaystyle\leqq\left\lvert a-x_{n}\right\rvert+\left\lvert x_{n}-b\right\rvert$ | (26.6) | ||

$\displaystyle=\epsilon+\epsilon$ | (26.7) | |||

$\displaystyle=2\epsilon$ | (26.8) |

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.