A question of numbers (one for the maths fans)

For the record, I learned the "proper" answer to that in a 1st year undergraduate course. But anyone with an understanding of recurring decimals ought to be able to work it out.

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From an ignorant view, it looks like anytime we have all these numbers in a row, our minds can't handle it, so we make hillarious rules to make the numbers fit into our minds.

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From a more educated point of view, the rules for summing infinite series make perfect sense. Our understanding if the integral calculus is based on them (and that's a very useful bit of mathematics).™

The math I am talking about (Real Analysis and the Theory of Limits) is basic undergraduate stuff. My first exposure to the Integral Calculus was before University. If your math skills are up to it then you should be able to find textbooks covering it.

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To me, if .999... can equal 1, then there is a limit to how many recurring 9's you can have

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That's exactly wrong. 0.999... = 1 because there is no limit to the number of '9's after the decimal place. That is how it works.For any finite number of '9's 1 - 0.999...9 = 0.000...1
For an infinite number of '9's 1 - 0.999... = 0.000...Because it's infinite there's no point to stop and put a '1'.
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One symbol looked like a funny looking E, what does that mean?

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I think what you are referring to is a capital sigma (Greek).
(Mathematcs also uses a backwards 'E' for "there exists").The sigma stands for "sum". In the wikipedia discussion there's an example with 'k = 0' below it and the 'infinity' symbol (like an '8' on it's side - above). What it means is "what you would get get if you worked out the formula for k = 0, k = 1, k = 2,... on to infinity and added all the results together".Since you can't actually do that you have to use other mathematics to work out what the result would be, More precisely what you work out is the limit of the sum as k tends to infinity (i.e. where it's going as you add on the results for higher and higher values of k - but it usually won't reach that value if you stop at any finite number of results)