@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 N_{0} 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 N**_{0} such that all integer m *greater than or equal to* 0 implies |c-E|>|c-w(N_{0}+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 N_{0} 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 x_{n} 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(N_{0}+m) replaces x_{n}

N_{0}=n_{0}

all whole m *greater than or equal to* 0 replaces n * greater than or equal to* n_{0}

The sequence of interest is w(n) instead of x_{n}.

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.