Lecture 10 - More on Sequences

Last time we talked about the idea of being pedantic with sequences (and consequently series). We needed to be careful about our definition of Convergence of a Sequence:

Convergence of a sequence

A Sequence $(a_{n})$ of real numbers converges to some $L \in R$ if:
$\forall ε > 0 \exists N \in N \forall n \geq N (| a_{n} - L | < ε)$

Geometrically this is saying:

Given some error $ε$ , you can always choose a start point in the sequence $a_{n}$ at point $N$ such that all further terms are within the range of $a_{n} - L$ and $a_{n} + L$ .

We write:
$L = lim_{n \to \infty} a_{n} = lim a_{n} \equiv a_{n} \to L$
when $L$ is the limit of $(a_{n})$ .

If $∄ L$ then we say $(a_{n})$ diverges; namely if $(a_{n})$ doesn't converge then it diverges.

Thus we get the Epsilon-Neighborhood for each limit $L$ (or $a$ for that note):

ε

-neighborhood of

a

$V_{ε} (a) = (a - ε, a + ε) = {x \in R | | x - a | < ε}$

More Notes on These

The definition of Convergence of a Sequence $a_{n} \to L$ involves nested quantifiers. Namely we have $\forall \dots \exists \dots$ where each of these quantifiers are being nested. For the definition for convergence we have:

(\forall ε > 0) (\exists N \in N) : n \geq N \Rightarrow | a_{n} - L | < ε

The important thing here is that you construct your $N$ based on $ε$ , so $N = f (ε)$ for some function on $ε$ . The idea here is that we really need to work backwards here; the smaller the $ε$ the bigger the $N$ (so you'll use some inverse relationship most likely).

Example

Let $a_{n} = \frac{n + 2}{n + 5}$ . Show that $a_{n} \to 1$ .

Proof
Let $ε > 0$ . This will always be your first statement.

Scratch

We need to magically pull out an $N$ out of our a** to be done. We have to do some scratch work here. We know that we want:

| a_{n} - 1 | < ε

Now try to work backwards to get a result of $n \geq N$ :

\begin{aligned} | a_{n} - 1 | & < ε \\ | \frac{n + 2}{n + 5} - 1 | & < ε \\ | \frac{- 3}{n + 5} | & < ε \\ \frac{3}{n + 5} < ε & (n + 5 > 0) \\ \frac{3}{ε} < n + 5 \\ \frac{3}{ε} - 5 < n \end{aligned}

But look! All we need is to construct some integer bigger than $\frac{3}{ε} - 5$ and we are done. As such, choose $N > \frac{3}{ε} - 5$ .

The Proof

Choose $N \in N$ such that $N > \frac{3}{ε} - 5$ (notice that $N, ε$ are inverses, which is a good sign). We can always choose this $N$ via the Cantor's Approach#^5404f9. Suppose that $n \geq N > \frac{3}{ε} - 5$ in this case:

\begin{aligned} | a_{n} - 1 | & = | \frac{n + 2}{n + 5} - 1 | \\ = | \frac{n + 2}{n + 5} - \frac{n + 5}{n + 5} | \\ = | \frac{- 3}{n + 5} | \\ \leq \frac{3}{N + 5} & n \geq N \\ < \frac{3}{(\frac{3}{ε} - 5) + 5} \\ = ε \end{aligned}

Thus $| a_{n} - 1 | < ε$ so we are done.
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We don't want to always prove that limits of specific sequences like our $a_{n}$ above have specific limits $L$ . Instead, we'll want to prove some theorems and laws so that we can use those instead of the limit proof directly.

For example see how we can prove that Limits are Unique:

Unique Limits

If a sequence has a limit, then the limit is unique.

Proof
Let $a_{n} \to a$ and $a_{n} \to b$ are both true. We want to show that $a = b$ . We know that then $\forall ε > 0 \exists N \in N (| a_{n} - a | < ε)$ and $ε > 0 \exists N \in N (| a_{n} - b | < ε)$ .

Here let $ε > 0$ (I know we aren't proving convergence, but we want to prove that $| a - b | < ε$ since then $a = b$ . If you're confused see Least Upper Bounds and Greatest Upper Bounds and namely the Suprema Lemma for context). Since $ε > 0$ , and thus $\frac{ε}{2} > 0$ then there are $N_{1} \in N$ such that for all $n \geq N_{1}$ that $| a_{n} - a | < \frac{ε}{2}$ (why not just $ε$ ? Later we'll have to add two $\varepsilons in the worst case, so then we need half of this to show that it's still possible). Similarly there is a $N_{2} \in N$ where for all $n \geq N_{2}$ then $| a_{n} - b | < \frac{ε}{2}$ .

Now with that then:
$\begin{aligned} | a - b | & = | (a - a_{n}) + (a_{n} - b) | \\ \leq | a - a_{n} | + | a_{n} - b | & Triangle Inequality \\ = | a_{n} - a | + | a_{n} - b | \\ < \frac{ε}{2} + \frac{ε}{2} \\ = ε \end{aligned}$
Thus $| a - b | < ε$ so then $a = b$ as required.
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