Tomorrow_Farewell

Now, ask yourself this question, 'is 0.999..., or any real number for that matter, a series?'. The answer to that question is 'no'.

You seem to be extremely confused, and think that the terms 'series' and 'the sum of a series' mean the same thing. They do not. 0.999... is the sum of the series 9/10+9/100+9/1000+..., and not a series itself.

EDIT: Also, the author does abuse the notations somewhat when she says '1+1/2+1/4 = 2' is a geometric series, as the geometric series 1+1/2+1/4+... does not equal 2, because a series is either just a formal sum, a sequence of its terms, or, in German math traditions, a sequence of its partial sums. It is the sum of the series 1+1/2+1/4+... that is equal to 2. The confusion is made worse by the fact that sums of series and the series themselves are often denoted in the same way. However, again, those are different things.
Would you mind providing a snippet with the definition of the term 'series' that she provides?

EDIT 2: Notably, that document has no theorem that is called 'convergence theorem' or 'the convergence theorem'. The only theorem that is present there is the one on convergence and divergence of geometric series.

Ok. In mathematical notation/context, it is more specific, as I outlined.

It is not. You will routinely find it used in cases where your explanation does not apply, such as to denote the contents of a matrix.

Furthermore, we can define real numbers without defining series. In such contexts, your explanation also doesn't work until we do defines series of rational numbers.

Ok. Never said 0.999... is not a real number

In which case it cannot converge to anything on account of it not being a function or any other things that can be said to converge.

because solving the equation it truly represents, a geometric series, results in 1

A series is not an equation.

This solution is obtained using what is called the convergence theorem

What theorem? I have never heard of 'the convergence theorem'.

0.424242... solved via the convergence theorem simply results in itself

What do you mean by 'solving' a real number?

0.999... does not again result in 0.999..., but results to 1

In what way does it not 'result in 0.999...' when 0.999... = 1?

You seem to not understand what decimals are, because while decimals (which are representations of real numbers) '0.999...' and '1' are different, they both refer to the same real number. We can use expressions '0.999...' and '1' interchangeably in the context of base 10. In other bases, we can easily also find similar pairs of digital representations that refer to the same numbers.

I meant what I said: "know patterns of repeating numbers after the decimal point."

What we have after the decimal point are digits. OTOH, sure, we can treat them as numbers, but still, this is not a common terminology. Furthermore, 'repeating number' is not a term in any sort of commonly-used terminology in this context.

The actual term that you were looking for is 'repeating decimal'.

Perhaps I should have also clarified known finite patterns to further emphasize the difference between rational and irrational numbers

No irrational number can be represented by a repeating decimal.

The explanation I've seen is that ... is notation for something that can be otherwise represented as sums of infinite series

The ellipsis notation generally refers to repetition of a pattern. Either ad infinitum, or up to some terminus. In this case we have a non-terminating decimal.

In the case of 0.999..., it can be shown to converge toward 1

0.999... is a real number, and not any object that can be said to converge. It is exactly 1.

So there you go, nothing gained from that other than seeing that 0.999... is distinct from other known patterns of repeating numbers after the decimal point

In what way is it distinct?
And what is a 'repeating number'? Did you mean 'repeating decimal'?

The decimals '0.999...' and '1' refer to the real numbers that are equivalence classes of Cauchy sequences of rational numbers (0.9, 0.99, 0.999,...) and (1, 1, 1,...) with respect to the relation R: (aRb) <=> (lim(a_n-b_n) as n->inf, where a_n and b_n are the nth elements of sequences a and b, respectively).

For a = (1, 1, 1,...) and b = (0.9, 0.99, 0.999,...) we have lim(a_n-b_n) as n->inf = lim(1-sum(9/10^k) for k from 1 to n) as n->inf = lim(1/10^n) as n->inf = 0. That means that (1, 1, 1,...)R(0.9, 0.99, 0.999,...), i.e. that these sequences belong to the same equivalence class of Cauchy sequences of rational numbers with respect to R. In other words, the decimals '0.999...' and '1' refer to the same real number. QED.

sponsorship by Anduril

Oof.

I'll try to find the time. However, I can't say I'm organised and willing enough to engage outside of socialist spaces to guarantee doing so.

IIRC, when I went into a TTY while on the lock screen, I was already working as the nixos user. Is that what you're asking?

IIRC in the live environment, the password for the "nixos" user is also "nixos"

That was the first thing I tried.

TTY1(?) and what I see elsewhere say that the password is empty.

Good bot.

I ran $ curl -s "https://archlinux.org/mirrorlist/?country=FR&country=GB&protocol=https&use_mirror_status=on" | sed -e 's/^#Server/Server/' -e '/^#/d' | rankmirrors -n 5 -. Where should I see an option to update to KDE 6.1?

Yeah, I was trying to do stuff like sudo install. I was not aware that it is used in combination with other commands like that.

I was wondering what I was doing wrong with sudo.