The Taylor Series for $\ln(x)$ is $$\ln(x)=\sum^{\infty}_{n=1}(-1)^{n+1}\frac{1}{n}(x-1)^n$$ for $|x-1|<1$.
If an invertable function $f(x)$ had a domain containing $(0,\infty)$ and had a range restricted so that $|f(x)-1|<1$ over that range and had a nice interplay with the natural log, then the Taylor Series of $\ln(x)$, $T(x)$, combined with that function to make something like $T(f(x))$, sould be equal to $\ln(x)$. An example for $f(x)$ would $g(x)=\displaystyle\lim_{a\to \infty}x^{\frac{1}{a}}$. For all positive $x$ as $a$ gets larger $g(x)\to 1$, which is within the bounds in which $T(x)=\ln(x)$. If we want $T(g(x))$ to equal $\ln(x)$, it would be useful to first consider $\ln(g(x))$, and how to change it to $\ln(x)$, or more simply how to change $\displaystyle\lim_{a\to \infty}\ln\left(x^{\frac{1}{a}}\right)$ to $\ln(x)$. By the properties of logarithms the solution is $\displaystyle\lim_{a\to \infty}a\cdot\ln\left(x^{\frac{1}{a}}\right)$. Combining this with our Taylor Series results in the following "expansion." $$\ln(x)=\lim_{a\to\infty}\sum^{\infty}_{n=1}(-1)^{n+1}\frac{a}{n}\left(x^{\frac{1}{a}}-1\right)^n$$ As far as I know the above is true for all $x>0$, although I personally don't have the knowledge to evaluated the limit at prove the equality. If I haven't made any mistakes could some one tell me how to make the proof more rigourus as well as why I don't see this in more places if it is correct?