Lecture 13 - Series

See Series, Convergence of A Series:

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 $s_{m} \leq t_{k}$ .