- It starts with a sequence, of course, which is just a function on the natural numbers, e.g.
\[
a_k=\left(\frac{1}{2}\right)^k
\]
We could just as well think of it as
\[
a(x)=\left(\frac{1}{2}\right)^x
\]
remembering that we're only interested in the domain of natural numbers.
This is the "Zeno sequence": steps to a wall, say. Depends on whether,
in fact, motion is actually possible. Suppose so.
-
Suppose I have the series -- that is a sum, S, like an integral --
adding up an infinite amount of stuff:
\[
S=\sum_{k=1}^\infty\frac{1}{2^k} =
\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots
\]
The terms represent "room steps", and I start from the back wall.
Suppose I have a sandwich, and want to hand it to someone outside in
the hallway, and that I have to be able to reach out to give it to them
(they can't stick their hand in).
Then I have to get close enough for my arm to go through the door
frame. How many terms $n$ will I need of the series to hand over the
sandwich?
- That's a partial sum, $S_n$
\[
S_n=\sum_{k=1}^n\frac{1}{2^k}
\]
a term in a new sequence of partial sums,
$\{S_k\}_{k=1}^\infty$, whose limit (if it exists and is
finite) will be the value of $S$.
- My arm length is the tolerance;
- $n$ is the number of terms $a_k$ I'll need to go to get
that close;
- and $S_n$ will be the distance I've traveled at the moment
I hand over the sandwich.
In this case, we can compute $S$ itself, of course ($S=1$ -- it's right
to the threshold of the door); but we're usually talking about
approximating more complex series (perhaps leveraging off a geometric
or p-series, whose errors we know how to bound).
So $|S-S_n|<{\textrm{arm's length}}$
means taking $n$ terms and getting to within range to hand over the
sandwich.
- Is that clear, and is anybody hungry?