Expand ln x in a series of powers of (x-1), and show

Expand $\ln x$ in a series of powers of $(x-1)$, and show for which $x$ the expansion is valid.

Explanation of Method

Writing $\ln x=\ln(1+y)$ with $y=x-1$ reduces this directly to the standard Maclaurin series for $\ln(1+y)$, already known to converge on $-1

Substitute $y=x-1$ to reduce to $\ln(1+y)$.
Quote the known Maclaurin series for $\ln(1+y)$ and its convergence interval.
Translate back to $x$, and confirm the endpoint values directly.

Let $y=x-1$: $\ln x=\ln(1+y)$.

$$ \ln(1+y) = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n y^n, \qquad -1

Substituting back $y=x-1$: $$ \ln x = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n(x-1)^n, \qquad -1

Property	Value
Series	$\sum(-1)^{n-1}(x-1)^n/n$
At $x=2$	Converges conditionally, sum $=\ln2$
At $x=0$	Diverges (undefined)
Region of validity	$0

This is literally Problem 228's series in disguise, recentered: $\ln(1+y)$ with $y=x-1$ is exactly the function studied there, so every conclusion transfers directly without recomputation.
The boundary of convergence at $x=0$ coincides exactly with the natural singularity of $\ln x$ itself — a case where the radius of convergence ($R=1$, centered at $x=1$) is forced entirely by where the function stops being defined, rather than by any subtler analytic obstruction.
Numerically at $x=3/2$: the series gives $\frac12-\frac18+\frac1{24}-\cdots\approx0.405$, matching $\ln(1.5)\approx0.405$ — confirming the expansion well inside its interval of validity.

$$ \ln x = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n(x-1)^n, \quad \text{valid for } 0