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See how to prove $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=1$

$x=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$

$2x=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$

Then:

$x=1$

Now I use the same argument to prove $2+4+8+...=-2$

$x=2+4+8+...$

$2x=4+8+16+...$

Then:

$x=-2$

Taha Akbari
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    The series you are writing down to represent $x$ does not converge, and subtracting divergent series (particularly, $\infty - \infty$) is in general undefined. – Nicholas Stull Jun 15 '16 at 10:10
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    The argument shows that if you have a way of making sense of the sum, and if that way satisfies certain basic properties, then the sum would have to be $-2$. But the value is conditional on you having a way of making sense of the sum. In the standard approach to sums of real numbers, the sum is just divergent, and your first line "$x=\ldots$" is just wrong. In fact, if you are in a framework where a sum of positive numbers must be positive, this is a proof by contradiction that the sum cannot exist. – Aaron Jun 15 '16 at 11:02
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    An infinite sum (aka a series) only has a value, if it converges. To decide whether a series converges you need to specify a metric. The tag real numbers tells me that you think of this as series of real numbers with metric determined by the usual absolute value. In that case the answer is simply that this series diverges, and has no value. Therefore the process of trying to determine a value is not on solid footing (basically because many steps ASSUME that the series has a value). However, if you use the 2-adic metric, then the scene becomes markedly different (see Achille Hui's post). – Jyrki Lahtonen Jun 15 '16 at 14:12
  • (cont'd) Because awareness of the dependence of the convergence on the choice of metric, and the existence of metrics (on the set of rational numbers) other than the usual absolute value require a level of mathematical sophistication that a typical asker here does not possess, I'm inclined to reclose this as duplicate. However, apparently regular users disagree on the best course of action. – Jyrki Lahtonen Jun 15 '16 at 14:15
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    @Jyrki Lahtonen: I voted to reopen when the close was due to lack of context - which IMHO applies to PSQ posts, not to posts about points of mathematical confusion (and this has at least some minimal context). If this is viewed as duplicate, then closing it for that reason would be OK with me. It is true that the 0.999 question is similar, but it has the separate well known confusion/problem that some students don't recognize 0.999... as a series in the first place. At least this question is explicitly about a series; I didn't see immediately from the title that is was a duplicate. – Carl Mummert Jun 15 '16 at 15:04
  • @JyrkiLahtonen I actually can't vote to close as duplicate again, my vote to close was part of the five who first closed the question... (only three users had voted as "no context", the other two as duplicate) Well, we'll see what happens. – Najib Idrissi Jun 15 '16 at 15:10
  • "but when I show this proof they have nothing to say." Is that so? I find this surprising. – Did Jun 15 '16 at 16:58
  • why it isn't duplicated. – Taha Akbari Jun 15 '16 at 17:56
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    $$\sum_{-\infty}^{\infty}x^k = 0$$ – Count Iblis Jun 15 '16 at 18:11
  • @CountIblis Losing the OP still a little more? – Did Jun 16 '16 at 04:59
  • @CarlMummert He probably doesn't know that 0.99999... is a series. –  Jun 16 '16 at 06:06

5 Answers5

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In the proof, you have begun by implicitly saying "Assume that the sequence converges, and call the limit $x$". For $1/2 + 1/4 + 1/8 + \cdots$, you obtain $x = 1$. For $2 + 4 + 8 + \cdots$, you obtain $x = -2$.

But that means you have only proved "If $2 + 4 + 8 + \cdots$ converges, then it converges to $-2$". This does not show that the series $2 + 4 + 8 + \cdots$ actually converges. That has to be established separately.

In the case of $1/2 + 1/4 + 1/8 + \cdots$, you can prove by induction that the partial sums are bounded by $1$, and then use the monotone convergence theorem to prove that the series converges. Until you prove the series converges, you don't actually know from your previous calculation that the limit of $1/2 + 1/4 + 1/8 + \cdots$ is $1$. You only know that if $1/2 + 1/4 + 1/8 + \cdots$ converges, then the limit is $1$.

Now, you can't prove the series $2 + 4 + 8 + \cdots$ converges, because it doesn't. But you can prove that if it did converge the limit would be $-2$ - which is still consistent with the series not converging. This is all that your calculation shows.

So the method you are using is fine, but you have to remember that it is relatively meaningless to try to compute the value of a series if the series doesn't converge. When you use the method in the question to compute the value of a series, you have to prove separately that the series converges.

Carl Mummert
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$\def\bR{\mathbb{R}}\def\bQ{\mathbb{Q}}$Well, $\sum_{k\geq1} 2^k$ converges to $-2$, but not in $\bR$, it does so in $\bQ_2$, the $2$-adic completion of $\bQ$. (In $\bR$, it obviously diverges.)

Let me prove that $\sum_{k\geq0} 2^k\to-1$, which is the same (just multiply by $2$). We have, for the partial sums, $A_n:=\sum_{k=0}^{n-1} 2^k= 2^n-1$ and then $A_n-(-1)=2^n$ and $2^n\to0$ in the $2$-adic sense.

So what you discovered is that series that are divergent in $\bR$ need not be divergent in other completions of $\bQ$.

yo'
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Hint the series converges only if $|r|<1$ so your second proof is wrong as $|r|=2$

Archis Welankar
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The proof does work if you interpret it correctly.

In the proof for $\frac12 + \frac14 + \cdots = 1$, you are using the fact in $\mathbb{R}$, the successive term $\frac{1}{2^n}$ is getting smaller and smaller as $n$ getting bigger and bigger. The partial sums $$\sum_{n=1}^{N} \frac{1}{2^n} = \frac{\frac12 - \frac{1}{2^{N+1}}}{1 - \frac12} = 1 - \frac{1}{2^N}$$ getting closer and closer to $1$ as $N$ increases. This means in $\mathbb{R}$, we have

$$\sum_{n=1}^\infty \frac{1}{2^n} = \lim_{N\to\infty} \sum_{n=1}^N \frac{1}{2^n} \text{ exists and equal to } 1$$

If you look at the other sum $2 + 4 + \cdots$, the successive term $2^n$ doesn't getting smaller and smaller as $n$ increases. The corresponding partial sums doesn't converge. In order for your argument to work, a prerequisite is the partial sum converges. When the partial sums fail to converge, the difference of two indeterminates is an indeterminate. Your argument will fail and you cannot conclude $2 + 4 + \cdots = -2$ when working within $\mathbb{R}$.

However, there is more than one way to extend $\mathbb{Q}$. For each prime number $p$, there is a beast $\mathbb{Q}_p$ called $p$-adic numbers. In particular, for $p = 2$, the successive terms $2^n$ does getting smaller and smaller as $n$ increases. If you work within the $2$-adic numbers $\mathbb{Q}_2$, you argument will work and the series does converge to $-2$ there.

achille hui
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You can't subtract divergent series.

user246836
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    Why not? These "fastest gun in the west"-type answers are getting tiring. – Najib Idrissi Jun 15 '16 at 11:23
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    @NajibIdrissi: If you are the person who downvoted it, then you're misusing the downvoting system. As far as answering OP's question matters, this is a perfectly correct answer. – user246836 Jun 15 '16 at 11:32
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    The hover-text for the downvote arrow says "This answer is not useful". I don't think this answer is useful, hence a downvote :) Compare your answer with e.g. Carl Mummert's answer. – Najib Idrissi Jun 15 '16 at 11:34
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    @NajibIdrissi: Then your definition of useful is flawed.This is a useful answer because it answers OP's question and tells him why what he has written is not a paradox in mathematics. + I don't think talking about $p$-adic numbers to someone who still doesn't fully understand basic laws of calculus is "useful", if not harmful. – user246836 Jun 15 '16 at 11:36
  • When they write $x = $, they are making an implicit assumption that the series converges, and of course it is OK to subtract convergent series. – Carl Mummert Jun 15 '16 at 12:56
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    @NajibIdrissi The answer is quite simple: It is simply not well-defined –  Jun 15 '16 at 13:19
  • how do you know it is infinity – Taha Akbari Jun 15 '16 at 15:46
  • @TahaAkbari: Look at the partial sums, $s_n$. $2^n \leq s_n$ for any $n$. Now let $n$ go to infinity. – user246836 Jun 15 '16 at 15:54
  • @TahaAkbari: I just read your profile and noticed that you're in 8th grade. What definition of infinity do you know? – user246836 Jun 15 '16 at 16:01
  • @TahaAkbari when you write $x = something$ for a series, you mean that the limit that the series is approaching is $something$. It means that you can add as many terms as you want but you will never exceed that limit. $$S = {a(r^n -1)\over (r-1)}$$. Now in a convergent series $r = 1/w$, $$S = {a(1/w^n -1)\over (1/w-1)}$$ , now limit both sides with $\lim_{n \to \infty}$, You will see that $1/w^n \to 0$, thus $$S = {a\over (1-1/w)}$$ $$=S = {a\over (1-r)}$$. Now do the same for a divergent series and you will see that $r^n = w^n$, which is unbounded. thus the sum, $S$ is unbounded (infinite). –  Jun 15 '16 at 18:49
  • @TahaAkbari I will assume that you don't calculus. So my advice, if you do what you did here when learning calculus then you would be in great trouble like me. Never treat $\infty$ as a number because it is not, you can't do $\infty- \infty = 0$. Consider this $w = \infty + 1$, which is same as $w = \infty$ now $w - \infty = \infty + 1 - \infty \Rightarrow 0 = 1$, which of course incorrect. –  Jun 15 '16 at 19:09