<s>[INST] $1=2$ | Continued fraction fallacy It's easy to check that for any natural $n$
$$\frac{n+1}{n}=\cfrac{1}{2-\cfrac{n+2}{n+1}}.$$
Now, 
$$1=\frac{1}{2-1}=\frac{1}{2-\cfrac{1}{2-1}}=\frac{1}{2-\cfrac{1}{2-\cfrac{1}{2-1}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-1}}}}=\ldots
=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\dots}}}}},$$
$$2=\cfrac{1}{2-\cfrac{3}{2}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{4}{3}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{5}{4}}}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{6}{5}}}}}=\ldots
=\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\ldots}}}}}.$$
Since the right hand sides are the same, hence $1=2$. 
 [/INST] 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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