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?


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.

  • $\begingroup$ Just as an additional question, are there any values of $x$ in $\ln(1-x)$ such that there is a $\pi$ somewhere imbedded inside? $\endgroup$ – Frank Feb 23 '17 at 18:19
  • $\begingroup$ @Frank I'm not exactly sure what you mean by embedded, could you explain more? $\endgroup$ – Yiyuan Lee Feb 24 '17 at 0:08
  • $\begingroup$ @Frank Any expression involving $\pi$ will do the trick, provided $|x| < 1$. So $x=1/\pi$, $x=\pi-3$, $x=\pi^2/9$, etc., would work. But maybe you're looking for a value of $x$ that doesn't involve $\pi$, but where the continued fraction has $\pi$ in it somehow? $\endgroup$ – Théophile Feb 24 '17 at 20:10

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,

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

  • $\begingroup$ Hello, welcome to Math.SE. Please see the MathJax tutorial for help formatting mathematics using LaTeX. It makes it much easier to read. $\endgroup$ – Trevor Gunn Apr 18 '17 at 22:27

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