If a series, such as

$$\sum _{n=1}^{\infty}\frac{1}{{n}^{2}}=\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\cdots ,$$has a value $s$, then the finite sums

$${s}_{N}=\sum _{n=1}^{N}\frac{1}{{n}^{2}}=\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\cdots +\frac{1}{{N}^{2}}$$that we can compute must eventually get close to $s$.

Let us call ${s}_{N}$ the
$N$th partial sum. For
the infinite series to converge to a value $s$ it is necessary that
the *sequence*
$\left({s}_{N}\right)$ formed from the partial sums
converges to some definite number, which is going to be the sum of the
infinite series. Thus we see that convergence of a series is a special
case of convergence of sequences. Therefore we start our discussion with
sequences and return to series later. In fact,
quite a bit later.

Definition.

A sequence is a list of real numbers
${a}_{1},{a}_{2},{a}_{3},\dots $
given in a definite order
and indexed by natural numbers. Sometimes we may start
sequences with ${a}_{0}$, sometimes with
${a}_{1}$, and sometimes with ${a}_{27}$. The starting
point doesn't really matter. What matters is tha ${a}_{n}$
is defined for all $n\in \mathbb{N}$ from this starting point
onwards. We will refer to the sequence as
the sequence $({a}_{n}{)}_{n=1,2,\dots}$

or just the sequence $\left({a}_{n}\right)$

when the starting point and
the choice of dummy variable $n$ are clear from context.
It is helpful to remember that $n$ is a dummy variable in
the sequence $\left({a}_{n}\right)$

and could equally well have been some other
letter (with the same meaning) as in
the sequence $\left({a}_{k}\right)$

.

Given a sequence of real numbers $\left({a}_{n}\right)$
we need to say what it means for ${a}_{n}$ to get close to a number $l$ as
$n$ gets large

. In other words, we need to define what it means for
the *limit of the sequence $\left({a}_{n}\right)$ to be $l$ as $n$ tends to
$\infty $
*.

- ${a}_{n}=n$; this is the sequence of numbers $1,2,3,4,\dots $ and seems to be going to infinity, and so doesn't seem to tend to any real number $l$.
- ${b}_{n}=-n$; this is $-1,-2,-3,-4,\dots $ and seems to be going to negative infinity, so also doesn't seem to tend to any real number $l$.
- ${c}_{n}={\left(-1\right)}^{n}n$; this is $-1,2,-3,4,-5,6,\dots $ and also doesn't seem to tend to anything. Worse, although the terms seem to be getting bigger the sequence does not seem to be going to plus or minus infinity either.
- ${d}_{n}={\left(-1\right)}^{n}$; this is $-1,1,-1,1,-1,1,\dots $. At least the numbers in this sequence don't get big. But they just alternate between $\mathrm{+1}$ and $-1$ so the sequence doesn't seem to tend to anything.
- ${e}_{n}=\frac{{\left(-1\right)}^{n}}{n}$; this is $-1,1/2,-1/3,1/4,-1/5,1/6,\dots $. and at last we have one that looks like it is going to zero! (But it is doing it in a strange way through positive and negative terms.)
- ${f}_{n}=\frac{1}{n}$; this is $1,1/2,1/3,1/4,1/5,1/6,\dots $. and is also going to zero, but it is decreasing everywhere and seems rather easier to understand.
- ${g}_{n}=-\frac{1}{n}$; this is $-1,-1/2,-1/3,-1/4,-1/5,-1/6,\dots $ and is increasing towards zero.
- ${h}_{n}=\frac{{\left(-1\right)}^{n}}{1000000000}+\frac{1}{n}$; this is approximately $1,1/2,1/3,1/4,1/5,1/6,\dots $, so if we just looked at the first million terms we might think it was going to zero. But since we can see the actual definition of the sequence we are not so easily fooled.

*We imagine a conversation between a customer and a salesperson.
The customer is thinking of buying a very expensive sequence
$\left({a}_{n}\right)$ which he is told converges to $l\in \mathbb{R}$.*

CUSTOMER: You've told me that ${a}_{n}\to l$ as $n\to \infty $. So if I need my values of ${a}_{n}$ to be within a half of $l$ (i.e., $\left|{a}_{n}-l\right|<\frac{1}{2}$) what do I have to do?

SALESPERSON: Oh that's easy. Any value of ${a}_{n}$ will do that provided $n\u2a7e57$.

CUSTOMER: And what if I need $\left|{a}_{n}-l\right|<\frac{1}{4}$?

SALESPERSON: You get that provided $n$ is at least 102.

CUSTOMER: And $\left|{a}_{n}-l\right|<\frac{1}{1000000}$?

SALESPERSON: *(Thinks for a few minutes)*
For all $n\u2a7e5271892743$ it will be true that $\left|{a}_{n}-l\right|<\frac{1}{1000000}$.

CUSTOMER: I'm going to use this sequence a lot, and I have a lot of other numbers $\epsilon $ for which I need $\left|{a}_{n}-l\right|<\epsilon $. I hope you don't mind, but I couldn't bring them all with me today to show you.

SALESPERSON: You'll need to tell me a little more. What sort of numbers are these $\epsilon $?

CUSTOMER: Oh sorry! Yes I should have said that these numbers $\epsilon $ are always positive real numbers. But they could be very small indeed.

SALESPERSON: *(Looks at the box the sequence is in.)*
In that case, I think this sequence is just what you need. Do you
see the guarantee that the manufacturers have written on it?
*(They both read the small print on the box.
It says: *

CUSTOMER: That's perfect. Just what I want.

The dialogue just given shows that there are a number of subtle issues
to do with convergence of sequences. First there is an error term

$\epsilon $.
This must be a *positive* real number since otherwise $\left|{a}_{n}-l\right|<\epsilon $
cannot possibly hold. Also $\left|{a}_{n}-l\right|\u2a7d0$ will hold only if
${a}_{n}=l$ and we are interested in sequences that converge to some number
but need not ever equal that number.
Given some such positive error term, the sequence must be very close to
$l$, in particular having a distance from $l$ that is less than the allowed error.
But it's important to remember that it is not enough that our sequence just does this
once or twice; it must be within the error for *all* values from some point onwards.
That is,
$\left|{a}_{n}-l\right|<\epsilon $ holds for any $n\u2a7eN$

where $N$ is some natural number, the choice of which is allowed to depend
on $\epsilon $. If we put this together we get the
official definition of convergence.

Definition.

The sequence $\left({a}_{n}\right)$
converges to $l\in \mathbb{R}$ as $n\to \infty $
if: **for all $\epsilon >0$ in the reals there is a natural number $N\in \mathbb{N}$
such that $\left|{a}_{n}-l\right|<\epsilon $ holds for all $n\u2a7eN$
**.

If we use the symbols
$\forall $ to mean for all

,
$\exists $ to mean there exists

,
and $\Rightarrow $ to mean implies

, we can write this as:
*the sequence $\left({a}_{n}\right)$ converges to $l\in \mathbb{R}$ as $n\to \infty $ if*

If that seems quite complicated to you, you are right. It *is*
quite complicated. But it really is the simplest possible definition (sigh).
I've tried to explain where it comes from and why. There will be further
motivation and reasons later. At this point, the best thing you can do is
to learn this definition—either by recalling the dialogue, or by any other
means. The symbolic form with $\forall $, $\exists $ and $\Rightarrow $
is the most useful one to learn, for reasons that will become apparent soon.

Convergence to $0$ is a little easier than convergence to an arbitrary $l\in \mathbb{R}$ since if $l=0$ then $\left|{a}_{n}-l\right|$ can be written $\left|{a}_{n}\right|$. So:

Definition.

The sequence $\left({a}_{n}\right)$
*converges to zero as $n\to \infty $ if*

Sequences that converge to zero are called null sequences.

If we check through the examples $\left({a}_{n}\right)$ to $\left({h}_{n}\right)$ given earlier, we should be able to check that our new definition agrees with our original intuition. I won't do this here, but it could be done in lectures. Note that if the customer had been sold the sequence $\left({h}_{n}\right)$ he would have been sold a dud! This is because, although the definition works for very small $\epsilon >0$ including $\frac{1}{1000000}$, it doesn't work for $\epsilon =\frac{1}{1000000000}$. But then, the manufacturers wouldn't have been able to honour their guarantee, so our customer would at least have been able to get his money back if he had been following this course.