Continued Fraction Identity Problem

Question

Question: What went wrong in my work to$$\ln(1-x)=-\cfrac x{1-\cfrac x{2+x-\cfrac {2x}{3+2x-\cfrac {3x}{4+3x-\ddots}}}}\tag{1}$$

I started with the expansion$$\begin{align*}\ln(1-x) & =-x-\dfrac {x^2}2-\dfrac {x^3}3-\dfrac {x^4}4-\&\text{c}.\\ & =-x\left\{1+\left(\dfrac x2\right)+\left(\dfrac x2\right)\left(\dfrac x{3/2}\right)+\&\text{c}.\right\}\tag{2}\end{align*}$$ And by Euler's Continued Fraction, $(2)$ can be rewritten into$$\begin{align*} & -x\left\{1+\left(\dfrac x2\right)+\left(\dfrac x2\right)\left(\dfrac x{3/2}\right)+\left(\dfrac x{2}\right)\left(\dfrac x{3/2}\right)\left(\dfrac x{4/3}\right)+\&\text c.\right\}\\ & =-\cfrac {x}{1-\cfrac x{2+x-\cfrac {2x}{3+2x-\ddots}}}\end{align*}\tag{3}$$ However, if we set $x=2$, then we have the LHS as $\ln(1-2)=\ln -1=\pi i$. The RHS becomes $$-\cfrac 2{1-\cfrac 2{4-\cfrac 4{7-\cfrac 6{10-\ddots}}}}=\pi i\tag{4}$$ But the LHS is real while the RHS is imaginary. What went wrong?

Answer 1

The Taylor expansion is valid only for $|x| < 1$.

There should have been a constraint somewhere in the question that corresponds to this.

Also, on an unrelated note, the $\ln$ of a negative number is multi-valued.

Answer 2

You went wrong with conversion from the series to the Euler CF. It was almost right, but the numeric numerator factor should be squared. That is,

$$ \ln (1-x) = -\cfrac{x}{1 - \cfrac{1^2x}{2 + x - \cfrac{2^2x}{3 + 2x - \cfrac{3^2x}{\ddots}}}}. $$