It's easy to check that for any natural $n$ $$\frac{n+1}{n}=\cfrac{1}{2-\cfrac{n+2}{n+1}}.$$


$$1=\frac{1}{2-1}=\frac{1}{2-\cfrac{1}{2-1}}=\frac{1}{2-\cfrac{1}{2-\cfrac{1}{2-1}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-1}}}}=\ldots =\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\dots}}}}},$$

$$2=\cfrac{1}{2-\cfrac{3}{2}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{4}{3}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{5}{4}}}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{6}{5}}}}}=\ldots =\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\ldots}}}}}.$$

Since the right hand sides are the same, hence $1=2$.

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    they arent the same. One right hand side is a continued fraction that converges to 2, one converges to 1. By using the dots, you cut away information. – CBenni Jun 11 '13 at 10:10
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    Just because $\frac{n}{n+1}$ tends to 1 as $n$ tends to $\infty$ doesn't mean that when you take the whole limit, it is equal to the first expression where you just have $2-1$ in the denominator of the repeating stuff. Also, the fact that you get $1=2$ should immediately tell you that you have went wrong somewhere. – Andrew D Jun 11 '13 at 10:25
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    @MherSafaryan no you dont. Dots imply a repitition of a given structure; They are not a rigorous mathematical notation. – CBenni Jun 11 '13 at 10:58
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    This is a good example of how continued fraction notation can be confusing, +1 (maybe you could make the title more helpful though...) – Matthew Towers Jun 11 '13 at 11:11
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    Is there an implied question here? E.g. "what's wrong with this fake proof"? – LarsH Jun 11 '13 at 14:38
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    Uh ya...what's the question? – nzifnab Jun 11 '13 at 16:27
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    This is not a question. – Christian Chapman Jun 12 '13 at 16:34
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    Your voice didn’t go up at the end of the sentence like in a real question. – k.stm Jul 16 '13 at 10:02

7 Answers7


A variant: note that $$\color{red}{\mathbf 1}=0+\color{red}{\mathbf 1}=0+0+\color{red}{\mathbf 1}=0+0+\cdots+0+\color{red}{\mathbf 1}=0+0+0+\cdots$$ and $$\color{green}{\mathbf 2}=0+\color{green}{\mathbf 2}=0+0+\color{green}{\mathbf 2}=0+0+\cdots+0+\color{green}{\mathbf 2}=0+0+0+\cdots$$ "Since the right hand sides are the same", this proves that $\color{red}{\mathbf 1}=\color{green}{\mathbf 2}$.

B. Mehta
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    Damn, this is a simple, straightforward and elegant as one can expect! +1 – DonAntonio Jun 11 '13 at 11:27
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    @DonAntonio I think it is invalid. It should be written as $$\Large\color{blue}{1}=0+0+0+\cdots+\Large\color{blue}{1}$$ and $$\Large\color{red}{2}=0+0+0+\cdots+\Large\color{red}{2}$$ – Tunk-Fey Apr 13 '14 at 09:40
  • @Tunk-Fey What are you talking about? What is "invalid"? Why a comment to DonAntonio? – Did Apr 13 '14 at 09:42
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    @Did I'm sorry DonAntonio, it's wrong tag since both name start with D. I mean the way you write $$0+0+0+\cdots$$doesn't imply that the $0$ will continue endless? Meanwhile, at the end of the $0$ series should be $2$. Maybe, I'm the only one here didn't get your point. – Tunk-Fey Apr 13 '14 at 09:59
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    @Tunk-Fey You seem to seriously think that some people on this page wish to prove that 1=2. Then indeed you might be missing the point... – Did Apr 13 '14 at 11:19
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    @Did Well, I just think instead of showing other fallacies of $1=2$, why didn't show a valid proof that the argument in the question is wrong. – Tunk-Fey Apr 13 '14 at 11:35
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    @Tunk-Fey Because explanations by analogy may be valuable (as a long tradition shows). But you still did not explain why, in your first comment, you saw fit to throw around the word "invalid"... – Did Apr 13 '14 at 11:38
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    @Did Indeed, I explained why it's invalid in my comment. The way you write $0+0+0+\cdots$, imply that the $0$ will continue endless but in fact it should be bounded by $1$ or $2$ at the end of the $0$ series. – Tunk-Fey Apr 13 '14 at 11:42
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    @Did OK, since I'm the first one to start this 'debate', I won't continue it any further. It seems we agree to disagree. **:)** – Tunk-Fey Apr 13 '14 at 12:00
  • This kind of species is actually seen, but off course those are wrong. For example proof of 1+2+・・=-1/12. – Takahiro Waki Jan 24 '17 at 04:02
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    @TakahiroWaki Bad example, of a different nature. – Did Jan 24 '17 at 08:09
  • @Did I claim something like david joyce's. That's difference between infinitely adding 0 or 1. https://www.quora.com/Whats-the-intuition-behind-the-equation-1+2+3+-cdots-tfrac-1-12. Also, you don't know the conclusion of this calculation. – Takahiro Waki Jan 24 '17 at 09:28
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    @TakahiroWaki OK, let me be brief: you are out of your depth here and are just scanning my answers to troll them (relatively inefficiently, but) because I previously mentioned that some of your posts were bogus, and your little self did not like that. Since this game of yours is going on for some time now, I suggest that you find somebody else to bother, or better yet, that you stop trolling people and start posting some valuable content on the site. – Did Jan 24 '17 at 16:35
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    @Tunk-Fey I believe you've missed the point here. The idea is get you to think "well, what's wrong with this $0+0+\dots$?" Once you figured that out, don't go back arguing the answerer, but rather, apply it to the question at hand. You ask for valid proofs why the argument in the question is wrong, and this is exactly what this answer invokes you to do *yourself*. – Simply Beautiful Art Jun 26 '17 at 19:20

Another example where dots are misleading: $$1= \frac{ 1 \cdot \color{blue}{2} \cdot \color{green}{3} \cdot \color{red}{4} \cdots}{ 2 \cdot \color{blue}{3} \cdot \color{green}{4} \cdot \color{red}{5} \cdots} \leq \frac{1}{2}$$

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    The dots are not misleading here, it is clear in both interpretations what the dots are replacing, unlike the question and @Didi's example. – jwg Jun 27 '13 at 08:25
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    Maybe just rephrasing @jwg's comment, I fail to see what explains the $=$ sign here. – Did Aug 13 '13 at 09:06
  • I didn't want to make my comment dismissive; in fact, I don't really understand the problem. – Seirios Aug 13 '13 at 09:48
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    Seirios: this fallacy is based on the order of taking limits (like many others). The crux of it is that $\lim\left(\frac{n!}{n!}\right)$ is not the same as $\frac{\lim(n!)}{\lim(n!)}$. Mine and @Did's comments are trying to say that this is not the same as 'hiding' two different expressions behind an ellipsis. – jwg Aug 13 '13 at 10:11
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    Would you prefer it if I deduced $1 \leq \frac{1}{2}$ from $1= \frac{1}{1}= \frac{1 \cdot 2}{1 \cdot 2} = \dots= \frac{1 \cdot 2 \cdot 3 \cdots}{1 \cdot 2 \cdot 3 \cdots}$ and $\frac{1}{2}= \frac{1}{1 \cdot 2} \geq \frac{1 \cdot 2}{1 \cdot 2 \cdot 3} \geq \frac{1 \cdot 2 \cdot 3}{1 \cdot 2 \cdot 3 \cdot 4} \geq \dots \geq \frac{1 \cdot 2 \cdot 3 \cdots}{1 \cdot 2 \cdot 3 \cdots}$? – Seirios Aug 13 '13 at 11:01
  • My point is that there is only one way to interpret $\frac{1\cdot2\cdot3\cdots}{1\cdot2\cdot3\cdots}$ in your comment, and it does not yield a result $\leqslant\frac12$. Likewise, there is only one way to interpret $\frac{1\cdot2\cdot3\cdots}{2\cdot3\cdot4\cdots}$ in your post, and it does not yield the result $1$. – Did Aug 13 '13 at 22:24

The first expression is a continued fraction, the second isn't. A continued fraction is the limit of $$ a_0, a_0 + \frac{1}{a_1}, a_0 + \frac{1}{a_1 + \frac{1}{a_2}} \ldots $$ for a fixed sequence of natural numbers $a_0, a_1, a_2 \ldots$

The second expression is the limit of fractions which look similar to these fractions, but which don't correspond to one well-defined sequence of naturals. The first lot of dots (between the equals signs) are ok, they just mean 'take the limit of this process'. This limit exists and is equal to 2, as you correctly deduce. The dots at the bottom of the final expression falsely suggest that the limit is a continued fraction, with coefficients given by the obvious sequence (eg $2, 2, 2, 2, \ldots$ implies the sequence consisting of only twos).

As @Did points out very elegantly, the same rules apply to infinite sums, and seem more obvious there - an infinite sum is not the same as the limit of an infinite sequence of sums, each with more terms in it than the one before. The common terms have to agree for any two sums.

I think this misunderstanding arises because sometime in iteration and limits we have the sense that the initial terms don't really matter, and sometimes this is the case. The first few terms of a sequence don't affect the limit, or the limit of the average, etc.

You are taking two different starting terms and iteratively applying a transformation to them. As you point out, this transformation doesn't actually change the number. This fact means though, that the starting value never becomes unimportant, and the final terms of each expression in your sequence similarly never become unimportant.

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This is a general feature of continued fractions. What you are doing is iterating the function $$ s(x) = \frac{1}{2-x} $$ ... so, for instance, you have that $$ 1 = s(1) = s(s(1)) = s(s(s(1))) = \cdots $$ and $$ 2 = s(3/2) = s(s(4/3)) = s(s(s(5/4))) = \cdots .$$

The function $s$ has a single fixed point at 1 (first line). It is known that iterating $s$ from any starting point except $x=2$ converges to 1 (think of $s$ as a mapping of the Riemann sphere; it takes $2 \mapsto \infty$ and $\infty \mapsto 0$). For $x<1$, this is easy to check: $x<s(x)<1$.

So, if $x\neq 2$, then $$ \lim_{n \to \infty} s^n(x) = 1 , $$ where $s^n(x)=s(s(\cdots s(x)))$ is the $n$th iterate of $s$.

But if you start with some $y$, it's always possible to find an $x_n$ so that $s^n(x_n) = y$ (including $\infty$, the map is one-to-one). You've shown that $$s^n\left(\frac{n+2}{n+1}\right) = 2.$$

Another nice thing to note here is that to end up with a limit above 1, we need the numbers at the end of the fractions (the $x_n$) to approach 1 from above: i.e., if for all $n$ that $$ s^n(x_n) = y > 1 $$ then $$ x_n \searrow 1 .$$

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  • What does the diagonal arrow mean in your last expression? Sorry if I am a bit amateur to this. – Mr Pie Mar 04 '18 at 00:00
  • Converges from above, i.e., $1 < \cdots < x_n < x_{n-1} < \cdots < x_1$. But: it's a bit wrong, it should say "if $s^n(x_n) = y > 1$ for all $n$ then $x_n \searrow 1$." – petrelharp Mar 05 '18 at 03:16
  • I would just write $x_n\to 1^+$. – Mr Pie Mar 05 '18 at 06:50

This is of the same type as the following

$$0=(1-1)+(1-1)+(1-1)+\ldots=1+(-1+1)+(-1+1)+(-1+1)+\ldots = 1 \; .$$

Did's example is even simpler and closer to yours in spirit.

In working with an infinite number of operations, you have to be very careful about how you are performing them. Normally, one uses some kind of limit, but then what you really do is define a sequence of finite but ever increasing number of operations. Changing something in the order of these operations will change the entire limit. Or in your case, you hide away the fact that in each term of your sequence, the last operation is subtracting a different number, $1$ in the first case, $(n+2)/(n+1)$ in the second.

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Today one friend of mine show a way to the fallacy $1=2$ in the given way described below.

Note that $$\begin{align} \log \,2 &=\log\, (1+1) \\&=1 -\dfrac 12 +\dfrac 13 -\dfrac 14 +\dfrac 15-\dfrac 16 +\dfrac 17 -\cdots \\ &= \left(1 +\dfrac 12 +\dfrac 13 +\dfrac 14 +\dfrac 15+\dfrac 16 +\dfrac 17 +\cdots\right)-2\times\left(\dfrac 12 +\dfrac 14 +\dfrac 16 +\dfrac 1{10}+\dfrac 1{12} +\dfrac 1{14} +\cdots\right)\\&=\left(1 +\dfrac 12 +\dfrac 13 +\dfrac 14 +\dfrac 15+\dfrac 16 +\dfrac 17 +\cdots\right)-\left(1 +\dfrac 12 +\dfrac 13 +\dfrac 14 +\dfrac 15+\dfrac 16 +\dfrac 17 +\cdots\right)\\&=\left(1-1\right)+\left(\dfrac12-\dfrac12\right)+\left(\dfrac13-\dfrac13\right)+\left(\dfrac14-\dfrac14\right)+\left(\dfrac15-\dfrac15\right)+\left(\dfrac16-\dfrac16\right)+ \cdots\\&=0+0+0+0+0+0+\cdots \\&=0\\&=\log\,1\end{align}$$ This implies $1=2.$ Hence it is proved.

That's all from me.

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    Another way: Clearly "That's all from me" stands for "That's all ... from me". If you take the limit from the left you get "That's all [initiated] from me", while if you take the limit from the right you get "That's all [terminated] from me", which clearly is not the same limit, as a translation to non-mathematical language gives "That's all invented by me" versus "That's all I have to say today." – PatrickT Mar 03 '18 at 07:01
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    @PatrickT I didn't realise this was an English lesson :) – Mr Pie Mar 04 '18 at 00:02

Incidentally, nobody appeared to have resolved the fallacy of the question, so I have provided an answer.

For all $a\in \mathbf N$, it follows $$\cfrac{1}{1+a}=1-\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\ddots - \cfrac 12}}}$$ such that the number of times the reciprocal in the continued fraction appears is $a$.

Proof. Note the identity $$\cfrac{1}{1+a}=1-\cfrac{1}{1+\color{red}{\cfrac 1a}}.$$ By letting $a=b-1$, it follows $$\cfrac 1b = 1-\cfrac{1}{1+\cfrac{1}{b-1}}.$$ From this we can substitute for $\color{red}{\cfrac 1a}$. $$\therefore \cfrac{1}{1+a}=1-\cfrac{1}{2-\cfrac{1}{1+\cfrac{1}{a-1}}}.$$ Clearly we can now likewise substitute for $1/(a-1)$, and the pattern will continue until for some $k\in\mathbf N$, the denominator of $1/(a-k)$ reaches $a-k=1$ since it cannot pass $0$. In consequence, we deduce as desired. (And, of course, when $a=0$, we have $1/1 = 1-0$.) This completes the proof. $\;\bigcirc$

And now, since $$\lim_{a\to\infty}\frac{1}{1+a}=0$$ then $$\boxed{\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\ddots}}}=1}$$

Mr Pie
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