Consider the geometric series with a = 1 and r = 1/10.
Then, we have
$$
\sum_{n=1}^{\infty}\left(\frac{1}{10}\right)^{n} = \sum_{n=0}^{\infty}9\left(\frac{1}{10}\right)^{n} – 9 = 9\left[ \lim_{k\to\infty}\left(\frac{ar^{k}}{r – 1}\right) + \frac{a}{1-r} – 1 \right] = 9\left[ \lim_{k\to\infty}\left(\frac{1\cdot(1/10)^{k}}{1/10 – 1}\right) + \frac{1}{1-1/10} – 1 \right] = 9\left(0 + \frac{10}{9} – 1 \right) = 9 \left( \frac{1}{9} \right) = 1.
$$
Hence, we say that the geometric series with a = 1 and r = 1/10 converges to 1.
Now, (correct me if I am wrong) the given series will never achieve the sum of 2 (rather, approach arbitrarily close) to 2. I conclude such because the limit of the quotient after the second equals sign never achieves its limit of 0.
Furthermore, there are many generally accepted ways to show that $0.\overline{9} = 1$; that is identically equal to 1.
One conventional way, of course, is to let $x=0.\overline{9}$ and subtract it from $10x$ giving $9x = 9$, and thus $x = 1$. And I've heard nobody complain about that—even though we must subtract an infinite number of terms in the exact same manner as we do a finite number of terms.
But, that means
$$
\underbrace{1 = 0.\overline{9} = \sum_{n=1}^{\infty}\left(\frac{1}{10}\right)^{n}}_{\text{exactly equal to}} = \overbrace{\sum_{n=1}^{\infty}\left(\frac{1}{10}\right)^{n} = 1}^{\text{arbitrarily close to}}.
$$
Doesn't it?
It seems to me that the only way the above equalities can hold is if converges to always means exactly equal to.
Otherwise, do we have a paradox here?
Please explain.
Thank you.
Best Answer
I'll try to explain some concepts super-explicitly, and you can say whether or not this clarifies your questions:
A series (or sequence) isn't a number. It's actually an infinite set of numbers (indexed by the positive integers). So a series is never equal to a number. A series has a limit (sometimes), and that limit is a number.
The phrase "the series converges to $a$" is just a different way of saying the limit of the series exists and is equal to the number $a$.
The phrase "the series gets arbitrarily close to $a$" is yet another way of saying that the limit exists and is equal to $a$, one that gestures towards the $\epsilon$ - $N$ definition of limit.
A fact about infinite sequences is that they can have a limit that is not equal to any of their terms. So "the given series will never achieve the sum of ..." isn't a problem.
The notation "$0.\overline{9}$" should be strictly thought of as a limit of the infinite sequence it describes. Notationally it works just like $\lim_{n \rightarrow \infty} ...$ or $\sum_{i=1}^\infty$.
Responding to comment:
1 - When you say "the limit of that sequence of partial sums merely approaches (but does not hit) its limit" you're going wrong.
The sequence of partial sums approaches a number.
The limit of the sequence of partial sums is a number.
2 - You're smart to be wary of proofs that manipulate infinite sequences (or any limits) just like finite arithmetical expressions. One can get into a lot of trouble that way. Many of those proofs that $0.\overline{9} = 1$ aren't rigorous and should be left behind -- let's cal them middle school proofs.
Firstly, theorems about manipulating infinite series usually start with "If certain limits exist, then ...." and the middle school proofs don't prove existence before doing the manipulations.
Secondly, remember that $0.\overline{9}$ (or, indeed, any non-terminating decimal) is notation for writing the limit of an infinite series. Limits of infinite series are too complicated for middle school, so we gloss over that distinction. But if you want to be really sure of what you're doing you have to deal with that.