@Castle
"A sequence is a series of numbers"
See edit.
"The second sequence is convergent, which means the total of the sequence converges to some number. In this case it's 1."
We don't sum the constituents of a sequence that way. {1/2, 1/4, 1/8, 1/16, 1/32, 1/64, ...} converges to 0. You mean to say {1/2, 1/2+1/4, 1/2+1/4+1/8, ...} converges to 1.
Edit: It looks (to me) like you quoted (in both of the above quotes) the first paragraph of Wikipedia page for series. What is in the first paragraph is wrong as a series is defined as the limit of a sequence of partial sums. The proper definition can be found here on Wolfram Mathworld.
"The way the "game" works is that one person has to find a number E that fits "between" the infinite sum and the target number, and the other person has to prove that it's impossible.
So in the case of our first sequence, you'd have to find a number E such that E is less than 1, but greater than any possible sum of the sequence. My job then is to find a number N that disproves you, because the sum of the first N terms in the sequence is at least E. Now doing this for discrete pairs of E/N would take forever, and leave open the possibility that there is some E that works and you weren't clever enough to find it. So I have to make a general solution that works for all possible Es you could pick, by turning the E automatically into an N that would disprove it."
That's not really how the game is played, our goal is for any E to find an N0 such that the following is satisfied:
We say {w(1), w(2), w(3), w(4), ...} converges to c if and only if for any chosen E there exists a positive integer N0 such that all integer m greater than or equal to 0 implies |c-E|>|c-w(N0+m)|.
If we can satisfy this statement it means we've succeeded in proving that there is a point in the sequence such that all points that follow it are closer to to c than E was. In simpler terms, it means that we've shown that the sequence eventually snuggles up next to c at a distance less than |c-E| and stays there. This snuggling is a cute way to say what it means to approach a limit, and is equivalent to Wolfram's epsilon-delta definition for a convergent sequence in one dimensional real standard metric space.* I took some liberty in changing how it was presented as it's easier to type and I find it easier to read on a text board like this.**
And if there is an E such that you can't find an N0 that satisfies the above statement, it either means the sequence doesn't converge or that c isn't the limit of w(n) as n gets large.
"We could use the epsilon-delta game. In this case, for any E you come up with, I just have to count the number of 9s it begins with, then add another one, and it will be between E and 1. Since there's no possible E between the sequence and 1, the sequence equals 1."
This statement is incorrect as it doesn't account for the distance |1-E| properly (as I could pick an E greater than 1).
*Sequences can be generalized quite a bit: We'd have to consider how we define convergent sequences in topological spaces (spaces that don't necessarily have a distance metric), and I'm not up to doing that right now. We run into all sorts of weird things, like if a topological space isn't Hausdorff then a sequence in the space can converge to more than one point, and can even converge to infinitely many points! Example.
**The exact original definition I learned in my classes was:
We say xn converges to c as n goes to infinity if and only if
image
Hooray for saving all those flash cards! Also some advice, in a professional setting use the original version I learned or what is on Wolfram, it looks better and they're the standard variants.
In the rewritten version I made the following major replacements to the TeX image directly above:
|c-E|=epsilon
w(N0+m) replaces xn
N0=n0
all whole m greater than or equal to 0 replaces n greater than or equal to n0
The sequence of interest is w(n) instead of xn.
And |a-b|=|b-a|, so the order there doesn't really matter.
Lastly, in addition to being equivalent to the version on Wolfram, the rewritten version is equivalent to the version I first learned as well.