On the Wikipedia article of the natural logarithm one finds two different series representations for $\ln(x)$:
- $\ln(x)= (x - 1) - \frac{(x-1) ^ 2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} \cdots$ for $x \in (0,2]$
- $\ln{x} = \sum_{n=1}^\infty {1 \over {n}} \left( {x - 1 \over x} \right)^n = \left( {x - 1 \over x} \right) + {1 \over 2} \left( {x - 1 \over x} \right)^2 + {1 \over 3} \left( {x - 1 \over x} \right)^3 + \cdots$ for $x \in [\tfrac 12,\infty)$
So on $x \in [\tfrac 12,2]$ one has
$$(x - 1) - \frac{(x-1) ^ 2}{2} + \frac{(x-1)^3}{3} \cdots = \left( {x - 1 \over x} \right) + {1 \over 2} \left( {x - 1 \over x} \right)^2 + {1 \over 3} \left( {x - 1 \over x} \right)^3 + \cdots$$
For $x \in (1,2)$ both series converge absolutely. So I thought, one can show the above equation for $x \in (1,2)$ by expanding the summands and reorder them. Unfortunately this does not seem to work. First question: Is there any easy way to show the equality of both series?
Second question: The right series also is
$$\left( {x - 1 \over x} \right) + {1 \over 2} \left( {x - 1 \over x} \right)^2 + {1 \over 3} \left( {x - 1 \over x} \right)^3 + \cdots = \left( 1 - {1 \over x} \right) + {1 \over 2} \left( 1 - {1 \over x} \right)^2 + {1 \over 3} \left( 1 - {1 \over x} \right)^3$$
and is thus a Laurent series after expanding all summands. The Laurent series does not seem to have any power $x^n$ with $n \ge 1$ (after expanding the summands there are just terms $\tfrac 1{x^n}$ with $n\ge 0$). How is it possible that such a sum leads to the equality with a Taylor series? The Taylor series does only have summands with the power $x^n$ ($n \ge 0$). Where do those summands come from?
