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Comment by tom-thistime | Hacker News Reader

Fine. Define 0.999... as the limit of the series sum(n=1 ... N)(10^-n), as N-> infinity. This is standard high school calculus. "Number" and "series" and "limit" and "convergence" don't all mean the same thing. However this number is defined as the limit of a convergent series. So the question really is meaningful. (One clue that this question is meaningful is the amount of space introductory calculus textbooks use to address it.)

Because I can still ask, in black and white, what law of "equality" do I use to establish that my limit equals 1? (It does, if I import the definition of "equality" from the real numbers. That's what they do in calculus class. )

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Thanks to a commenter who pointed out that my sum above should be

sum(n=1 ... N)(9*10^-n).

I can't, uh ... fully endorse that comment, which is not entirely accurate and doesn't answer my question. But I sure did miss the '9'.

You asked for a "law" of equality (whatever that is) and provided an answer that proves it converges to 1. What more could you possibly want?

We seem to agree on this: you don't think there's any need for a way to determine whether two real numbers are equal.

For ordinary math, though, using some criterion for equality (for example x>=y and y>=x) is basic and not controversial. So it seems unconvincing (to me) when you seem to imply the opposite.

> you don't think there's any need for a way to determine whether two real numbers are equal.

You are putting words in my mouth.

And you clearly do not understand the answer.

I guess I'm not very good at ELI5 because I very clearly answered your question with your own proposal.

Maybe when you get to college a professor can do a better job explaining it to you (if you actually make it to college, because you're going to struggle very hard if that's how you think when an answer is spoon-fed to you).

There is an easy way to prove two numbers are equal. Typically in the reals there are three possibilities: a > b, a < b, a = b. If you eliminate a > b and a < b then you are left to conclude a = b. And this is exactly what is done in Apostol's Calculus Vol 1 (IMHO the greatest calc book ever written) chapter 1 when he proves that the area under n^2 is EXACTLY (n^3)/3, with no "calculus". You would be shocked how far into calculus the author gets with just that theorem. Can't recommend that book enough.

Thanks, I'll take a look. I like that kind of thing very much.

I use applied math. I haven't taken a class in real analysis. But it's fun how often grinding out the solution to a "real world," practical PDE turns out not to actually be the nicest (simplest and/or clearest and/or sufficiently insight-producing) way to understand the (hopefully) corresponding physical problem in the lab.

Stripping off the "calculus" and replacing it by limits sometimes seems to help highlight alternate perspectives that the magic "integrals" and "derivatives" kind of conceal.

Even when it's not more effective, it's definitely more fun.

I'm not sure if you are asking for an answer or a rhetorical question? I'll assume the former.

Your terms are bit jumbled, so let's keep it simple: you're asking how to prove if an infinite sum converges and what its value is. Convergence proofs require analytic thought: meaning there may not be an immediate look-up. You need to convert the problem into the known corpus of convergent sums or use one of many tests (bounds test, integral test, etc) to show it converges analytically. Which you only learn through experience and memorization (unless you want to re-prove hundreds of series... maybe you do!) Fortunately this one is easily re-written as a known convergent sum.

First, you missed a term in your sum (9), re-written here:

sum(n=1..inf) 9 * 10^-n

Step 1: you pull out the 9 and it becomes 1/10+1/100+1/1000...

Step 2: Then we shift to n=0 by subtracting 1/10^0 from the series so that it is in the form n=0..k-1

1/10^0 + 1/10 + 1/100 + 1/1000 + ... + 1/10^-n - 1/10^0

Step 3: Now we've got ourselves a geometric series of just 1/10^n .. wikipedia does a great job explaining the sum convergence for GS from n=0...inf: https://en.wikipedia.org/wiki/Geometric_series

Step 4: compute geometric convergence

(1-r^n)/(1-r) = (1-(1/10)^n)/(1-1/10) = 1/(1-1/10) = 10/9

So we have 10/9 as the solution to Sum[n=0...inf](1/10^n)

Step 5: the remaining arithmetic

Now subtract our 1/10^0 ... and then * 9 = 1

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