Does ``Converge to'' and ``Strict Equality'' Always Mean the Same Thing? If Not, is This a Paradox?

Question

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.

Answer 1

Your doubts about: Does Converge to and Strict Equality Always Mean the Same Thing? might look quite reasonable at first sight.

When we say something like: Let $(a_n)_{n\geq 0}$ be a sequence of real numbers and $a\in\mathbb{R}$ so that \begin{align*} \lim_{n\to \infty}a_n=a\tag{1} \end{align*} Is it really so that the limiting value represented by the left-hand side of (1) is equal to $a$, so that the equality sign $=$ is justified?

We know from definition of the limit of a convergent sequence: For each $\varepsilon>0$ there is an index $N\in\mathbb{N}$, so that for all $n>N$ we have \begin{align*} |a_n-a|<\varepsilon \tag{2} \end{align*} and from that we conclude equality. But this is not obvious. In fact we need to recall an axiom in order to do so.

Here we use behind the curtain the axiom of trichotomy which states that each real number is either greater than, or equal to or less than zero.

So, in fact we axiomatically cope with the situation. However small we choose $\varepsilon>0$ we can find an index $N\in\mathbb{N}$ so that $|a_n-a|<\varepsilon$ for all $n>N$. But this means that the limiting value is neither greater than nor smaller than $a$. Thanks to the axiom of trichotomy we can conclude that equality holds.

Answer 2

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 $\sum_{i=1}^\infty$.

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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.

Answer 3

Now, (correct me if I am wrong) the given series will never achieve the sum of 2 (rather, approach arbitrarily close) to $1$.

You are .... not correct.

The series will never achieve $1$ within a finite number of terms and we reach a precise value that is close to $2$ depending upon the precise number of steps and we can achieve a result that is as arbitrarily close to $1$ as we like by performing a specific finite number of steps required to get it that close to $1$.

But if we do the series an infinite number of steps..... well, that's a different story altogether.

... Okay, we can debate what it means to do an infinite number of terms and if that even makes sense and if it does make sense is it possible.

But if doing it an infinite number of steps is meaningful (in my opinion, it isn't) and if figuring out the infinite sum is possible (in my opinion as it isn't meaningful it's not meaningful to talk of it being done) then, if so, then the sum is exactly $1$ and we do get $1$. Not arbitrarily close to $1$ (we aren't doing this a finite number of times and not getting to the end; we are doing this an infinite number of times and we ARE getting to the end.

More formally we have a sequence of terms:

$a_n = \sum_{k=1}^n 9\cdot \frac1{10^k}=0.\underbrace{999....9}_{n\text {times}}$

This sequence $\{a_i\} = \{0.9,0.99,0.999,0.9999,......\}$ converges to $1$ but none of the terms are equal to one but it ISN'T any of the terms we are talking about when we say the limit is exactly equal to $1$. (It is.) It's the limit of the terms, not any of the terms itself.

And we define an INFINITE sequence, not equal to any of the finite sequences $\sum_{k=1}^\infty 9\cdot \frac1{10^k} =\lim_{n\to \infty} \sum_{k=1}^n 9\cdot \frac1{10^k}$ and it is the limit that is the infinite series, that is not any of the finite terms that make the sequence, which is equal to exactly $1$.