An alternate power series representation for ln(x)

[bo198214]

\( \ln(x) = a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ne^{an}}(x-e^a)^n \)

But this follows directly from the definition of the logarithm, by the following equivalent transformations:
\( \begin{eqnarray}
\ln(x) &=& a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ne^{an}}(x-e^a)^n\\
&=& a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x/e^a-1)^n\\
\ln(e^a y) &=& a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(y-1)^n\\
a + \ln(y) &=& a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(y-1)^n\\
\ln(y) &=& \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(y-1)^n
\end{eqnarray} \)

sadly, doesn't converge for values less than 1

The \( \ln \) series converges for \( |y-1|<1 \), hence your series converges for \( |x/e^a-1|<1 \), this should include all values \( x\in (0,2e^a) \)?