Q.The expansion $\log(1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \ldots$ is valid for:

Recognising the series as a scaled logarithm series gives the sum $\dfrac32\log\left(\dfrac53\right)$.

Recall the standard expansion $\ln(1+x)=x-\dfrac{x^2}2+\dfrac{x^3}3-\dfrac{x^4}4+\ldots$ for $|x|<1$.

The given series is $1-\dfrac12\left(\dfrac23\right)+\dfrac13\left(\dfrac23\right)^2-\dfrac14\left(\dfrac23\right)^3+\ldots=\sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{n}\left(\dfrac23\right)^{n-1}$.

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The Maclaurin/logarithmic series for $\log(1+x)$ converges for $-1 < x \le 1$; it diverges at $x=-1$ (harmonic series) and is undefined for $x < -1$.

The series $\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$ is obtained by integrating the geometric series for $\frac{1}{1+x}$, which itself converges for $|x|<1$.

At the boundary $x=1$: the series becomes the alternating harmonic series $1 - \frac12+\frac13-\ldots$, which converges (to $\log 2$), so $x=1$ is included.

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