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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    $\begingroup$ 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. $\endgroup$ – CBenni Jun 11 '13 at 10:10
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    $\begingroup$ 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. $\endgroup$ – Andrew D Jun 11 '13 at 10:25
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    $\begingroup$ @MherSafaryan no you dont. Dots imply a repitition of a given structure; They are not a rigorous mathematical notation. $\endgroup$ – CBenni Jun 11 '13 at 10:58
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    $\begingroup$ Is there an implied question here? E.g. "what's wrong with this fake proof"? $\endgroup$ – LarsH Jun 11 '13 at 14:38
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    $\begingroup$ Uh ya...what's the question? $\endgroup$ – nzifnab Jun 11 '13 at 16:27

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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    $\begingroup$ The dots are not misleading here, it is clear in both interpretations what the dots are replacing, unlike the question and @Didi's example. $\endgroup$ – jwg Jun 27 '13 at 8:25
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    $\begingroup$ Maybe just rephrasing @jwg's comment, I fail to see what explains the $=$ sign here. $\endgroup$ – Did Aug 13 '13 at 9:06
  • $\begingroup$ I didn't want to make my comment dismissive; in fact, I don't really understand the problem. $\endgroup$ – Seirios Aug 13 '13 at 9:48
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    $\begingroup$ 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. $\endgroup$ – jwg Aug 13 '13 at 10:11
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    $\begingroup$ 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 $\frac{1}{2}= \frac{1}{1 \cdot 2} \geq \frac{1 \cdot 2}{1 \cdot 2 \cdot 3} \geq \frac{1 \cdot 2 \cdot 3}{1 \cdot 2 \cdot 3 \cdot 4} \geq \dots \geq \frac{1 \cdot 2 \cdot 3 \cdots}{1 \cdot 2 \cdot 3 \cdots}$? $\endgroup$ – Seirios Aug 13 '13 at 11:01

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 $\color{red}{\mathbf 1}=\color{green}{\mathbf 2}$.

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    $\begingroup$ Damn, this is a simple, straightforward and elegant as one can expect! +1 $\endgroup$ – DonAntonio Jun 11 '13 at 11:27
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    $\begingroup$ @DonAntonio I think it is invalid. It should be written as $$\Large\color{blue}{1}=0+0+0+\cdots+\Large\color{blue}{1}$$ and $$\Large\color{red}{2}=0+0+0+\cdots+\Large\color{red}{2}$$ $\endgroup$ – Tunk-Fey Apr 13 '14 at 9:40
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    $\begingroup$ @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. $\endgroup$ – Tunk-Fey Apr 13 '14 at 11:35
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    $\begingroup$ @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"... $\endgroup$ – Did Apr 13 '14 at 11:38
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    $\begingroup$ @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. :) $\endgroup$ – Tunk-Fey Apr 13 '14 at 12:00

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.


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 $$ x_n \searrow 1 .$$

  • $\begingroup$ What does the diagonal arrow mean in your last expression? Sorry if I am a bit amateur to this. $\endgroup$ – Mr Pie Mar 4 '18 at 0:00
  • $\begingroup$ 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 $x_n \searrow 1$." $\endgroup$ – petrelharp Mar 5 '18 at 3:16
  • $\begingroup$ I would just write $x_n\to 1^+$. $\endgroup$ – Mr Pie Mar 5 '18 at 6:50

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.


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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    $\begingroup$ 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." $\endgroup$ – PatrickT Mar 3 '18 at 7:01
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    $\begingroup$ @PatrickT I didn't realise this was an English lesson :) $\endgroup$ – Mr Pie Mar 4 '18 at 0:02

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 $$\boxed{\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\ddots}}}=1}$$


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