# Why are there two series representations of the natural logarithm?

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?

• One is an expansion in terms of ${1 \over x}$ and the other in terms of $x$. Note that $\log({1 \over x}) = - \log x$. Dec 8, 2014 at 20:00