Lecture 9 - Introduction to Sequences and Series

This will not be on the exam tomorrow, and this introduction will be pretty light.

First let's define a Sequence:

These are just ordered lists of elements. Essentially we talked about Countable Sets, so in the same vein we can have an order to the elements $a_{1}, a_{2}, a_{3}, . . .$ .

Sequence

A sequence is a function $f$ where the domain is just $N$ .

For instance, consider the sequence of numbers ${\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, . . .}$ . What do we mean that the sequence is a function with domain $N$ ? Essentially $1 \mapsto \frac{1}{2}, 2 \mapsto \frac{1}{4}$ , and so on. In general $n \mapsto \frac{1}{2^{n}}$ .

Logically, having either a finite sequence or an infinite sequence, both are countable, which is the key here.

Notation

We write a sequence in any of the following notations:

$(a_{1}, a_{2}, a_{3}, . . .)$
$(a_{n})_{n = 1}^{\infty}$
$(a_{n})_{n \in N}$
$(a_{n})$

From calculus ...

We may start with an index $n$ starting from $0$ or some other integer $k \in Z$ .

Why do we care about sequences? They're important for describing infinite sums, known as Series, Convergence of A Series.

Examples of Sequences

i. $(1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots)$
ii. $(\frac{1 + n}{n})_{n = 1}^{\infty}$
iii. $(a_{n})$ where $a_{n} = 2^{n}$ for each $n \in N$
iv. $(x_{n})$ where $x_{1} = 2$ and $x_{n + 1} = \frac{x_{n} + 1}{2}$

Before our definition, consider an infinite sum $\sum_{n = 1}^{\infty} a_{n} = a_{1} + a_{2} + a_{3} + \dots$ . Then the sequence of partial sums $s_{n}$ is just the first $n$ -terms summed up:
$s_{n} = \sum_{k = 1}^{n} a_{k} = a_{1} + a_{2} + \dots + a_{n}$
So given the definition of a convergent sequence, we'll have a definition for a convergent series as well.

Series

An infinite series is a formal expression of the form:
$\sum_{n = 1}^{\infty} a_{n} = a_{1} + a_{2} + \dots$
To this we associate a sequence of partial sums is the sequence $(s_{m})_{m \in N}$ defined as:
$s_{m} = a_{1} + a_{2} + \dots + a_{m}$
We say that $\sum_{n = 1}^{\infty} a_{n}$ converges to $s \in R$ and write:
$\sum_{n = 1}^{\infty} a_{n} = s$
To mean that $(s_{m}) \to s$ .

The difficulty here is that $(s_{m})$ is usually very hard to compute. But let's look at a simple one.

Example: Geometric Series

Let $r \in R$ . Consider:
$\sum_{n = 0}^{\infty} r^{n} = 1 + r + r^{2} + \dots$
$s_{m} = 1 + r + r^{2} + \dots + r^{m}$
Recall that in general that: $a^{k} - b^{k} = (a - b) (a^{k} + a^{k - 1} b + \dots + b^{k})$ (multiply out the right side to verify). With that then:
$1 - r^{m + 1} = (1 - r) (1 + r + r^{2} + \dots + r^{m}) = (1 - r) s_{m}$
So long as $r \neq 1$ (which diverges anyways) then:
$\Rightarrow s_{m} = \frac{1 - r^{m + 1}}{1 - r} = \frac{1}{1 - r} + (\frac{- 1}{1 - r}) r^{m + 1}$
Then from our talk on Monotonic Sequences:

Monotonic Sequences

A Sequence $(a_{n})$ is

increasing if $\forall n \in N$ we have $a_{n} \leq a_{n + 1}$
decreasing if $\forall n \in N$ we have $a_{n} \geq a_{n + 1}$
monotone if it is either increasing or decreasing.

Using the ALT

Using the Limit Laws (Algebraic Limit Theorem) then:
$\sum_{n = 0}^{\infty} r^{n}$
is:

converges $\frac{1}{1 - r}$ if $| r | < 1$
diverges otherwise

Showing any series converges.

Consider for a moment the series $\sum_{n = 1}^{\infty} a_{n}$ where $\forall n \in N$ , $a_{n} \geq 0$ and $a_{n + 1} \leq a_{n}$ . As a result, we have to have $(s_{m})$ be increasing (since there's no negative terms to decrease the partial sum), and thus monotonic. Using the Monotonic Sequences then the only way $\sum_{n = 1}^{\infty} a_{n}$ converges iff $(s_{m})$ is bounded above (obviously it is bounded below already).
So the name of the game to show the series converges is to show an upper bound for the partial sums. Similarly, to show it diverges, show that no such upper bound can exist.

Series converges

\equiv

Bounded Partial Sums

A series $\sum_{n = 1}^{\infty}$ where $a_{n} \geq 0$ and $a_{n + 1} \leq a_{n}$ for all $n \in N$ converges iff $(s_{m})$ , the sequence of partial sums, is bounded above.

Proof
See our remark above.
☐

Example 2

Consider the series:
$\sum_{n = 1}^{\infty} = 1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots$
It's a $p$ -series with $p = 2 > 1$ so we expect it to converge. But let's show it converges. We'll show that there's a bound for $s_{m}$ . Notice that:
$s_{m} = 1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots + \frac{1}{m^{2}}$
Notice that:
$\begin{aligned} s_{m} & = 1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots + \frac{1}{m^{2}} & Incomputable \\ \leq 1 + \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{(m - 1) m} & Slightly bigger, but computable \\ = 1 + (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{m - 1} - \frac{1}{m}) & \frac{1}{(x - 1) x} = \frac{1}{x - 1} - \frac{1}{x} \\ = 1 + 1 - (\frac{1}{2} - \frac{1}{2}) - (\frac{1}{3} - \frac{1}{3}) + \dots + (\frac{1}{m - 1} - \frac{1}{m - 1}) - \frac{1}{m} \\ = 2 - \frac{1}{m} \\ < 2 \end{aligned}$
Thus then $s_{m}$ is bounded above by $2$ , so then the series converges. Not only that, but by the Limits and Order (Order Limit Theorem), then it converges to some value less than 2.

Example 3: Harmonic Series

Consider:
$\sum_{n = 1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \dots$
we know that this diverges, but let's prove it. We have to show that $(s_{m})$ is unbounded. Then:
$\begin{aligned} s_{m} & = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{m} & Incomputable \\ = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{2^{k}} & Let m = 2^{k} for a moment \\ s_{2^{k}} & = 1 + \frac{1}{2} + \underset{\geq 2 \cdot \frac{1}{4}}{\underset{⏟}{(\frac{1}{3} + \frac{1}{4})}} + \underset{\geq 4 \cdot \frac{1}{8}}{\underset{⏟}{(\frac{1}{5} + \dots + \frac{1}{8})}} + \dots + \underset{\geq 2^{k - 1} \cdot \frac{1}{2^{k}}}{\underset{⏟}{(\frac{1}{2^{k - 1} + 1} + \dots + \frac{1}{2^{k}})}} \\ \geq 1 + \underset{k times}{\underset{⏟}{\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \dots + \frac{1}{2}}} \\ = 1 + \frac{k}{2} \end{aligned}$
Clearly as $k \to \infty$ then $1 + \frac{k}{2} \to \infty$ (diverges) so then $s_{2^{k}}$ must also diverge since it's unbounded.

Example 4: The General Case

Now consider the abstract, general case of $\forall n \in N$ where $a_{n} \geq 0$ and $a_{n + 1} \leq a_{n}$ . Let $s_{m} = \sum_{n = 1}^{m} a_{n}$ . Then let $m = 2^{k}$ similar to our previous example:
$\begin{aligned} s_{2^{k}} & = a_{1} + a_{2} + (a_{3} + a_{4}) + (a_{5} + \dots + a_{8}) + \dots + (a_{2^{k - 1} + 1} + \dots + a_{2^{k}}) \\ \geq a_{1} + a_{2} + 2 a_{4} + 4 a_{8} + \dots + 2^{k - 1} a_{2^{k}} \\ \geq \frac{1}{2} \underset{:= t_{k}}{\underset{⏟}{(a_{1} + 2 a_{2} + 4 a_{4} + \dots + 2^{k} a_{2^{k}})}} \end{aligned}$
So if $m \geq 2^{k}$ then $s_{m} \geq \frac{1}{2} t_{k}$ . Showing that $t_{k}$ is unbounded would imply that $s^{2^{k}}$ is unbounded, showing a divergent series.

We could do a bound in the other direction too:
$\begin{aligned} s_{2^{k}} & = a_{1} + \underset{\leq 2 a_{2}}{\underset{⏟}{(a_{2} + a_{3})}} + \underset{\leq 4 a_{4}}{\underset{⏟}{(a_{4} + \dots + a_{7})}} + \underset{\leq 8 a_{8}}{\underset{⏟}{(a_{8} + \dots + a_{15})}} + \dots + \underset{\leq 2^{k - 1} a_{2^{k - 1}}}{\underset{⏟}{(\frac{1}{a_{2^{k - 1}}} + \dots + \frac{1}{a_{2^{k} - 1}}) + a^{2^{k}}}} \\ \leq a_{1} + 2 a_{2} + 4 a_{4} + 8 a_{8} + \dots + 2^{k - 1} a_{2^{k - 1}} + 2^{k} a_{k} \\ = t_{k} \end{aligned}$
So if $m \leq 2^{k}$ then $s_{m} \leq t_{k}$ .

There are some cautionary tales that will help develop more rigorous definitions here.

Cautionary Tales

The First Tale

Consider the alternating harmonic series:

1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dots

from calculus we can use the AST to show that this converges. Specifically it is given that it converges to $\ln (2)$ , but this isn't important so say it converges to $s \in R$ :

s = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dots

Now multiply the series by $\frac{1}{2}$ :

\frac{1}{2} s = \frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \dots

Now add the halves from the $s$ equation with either other with the $\frac{1}{2}$ in the $\frac{1}{2} s$ equation:

\frac{3}{2} s = 1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \dots

is a rearrangement of the terms of the AHS. It seems that rearranging the series gives a different number.

Summary

The order of added terms dictates what number the series converges to.

Because $P (N)$ is uncountable, then we could reorder this in an uncountable order. This suggests that we could order these in any way, such that we could get any real number out. In fact, we claim that:

Claim

Given any $r \in R$ , there is some rearrangement of the AHS such that it adds up to $r$ .

This is crazy! This seems to break commutativity that we know in the finite sense.

The Second Tale

What if we do double summations (similar to that of doing double integrals)? Consider the series:

\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots = 1

Let's make it add to zero by doing a $- 1$ out front:

- 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots = 0

Remark that putting a $0$ out front doesn't change the sum:

0 - 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots = 0

and we can keep adding zeroes:

0 + 0 - 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots = 0

Putting these in a giant grid:

-1	1/2	1/4	1/8	1/16	...
0	-1	1/2	1/4	1/8	...
0	0	-1	1/2	1/4	...
0	0	0	-1	1/2	...
...	...	...	...	...	...
Let $a_{m n}$ denote the entry in the $m$ -th row and $n$ -th column. What is $\sum_{m \in N, n \in N} a_{m n}$ ?
If we go rows first then they all add to 0, by our arguments above. For the column first, we have:

1st column: -1
2nd column: -1/2
3rd column: -1/4
...
$n$ -th column: $\frac{- 1}{2^{n}}$

If we add all the values in the rows first, we expect $0$ since $0 + 0 + \dots = 0$ . So $Σ = 0$ right?If we add columns first instead we get:

- 1 - \frac{1}{2} - \frac{1}{4} - \dots = - 2 = Σ

So what is it? Is the series 0 or -2?

The Third Tale

Consider the series:

1 - 1 + 1 - 1 + 1 - 1 + \dots

We could add in groups like:

(1 - 1) + (1 - 1) + (1 - 1) + \dots = 0 + 0 + 0 + \dots = 0

But we could also group like:

1 + (- 1 + 1) + (- 1 + 1) + \dots = 1 + 0 + 0 + \dots = 1

so again which is it?

The Conclusion

The moral of these tales is that we need to be precise about what it means to converge! As such then

This allows us to define 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.