It is claimed that
$$ \ln(1+x) = \sum_{k=0}^{\infty}(-1)^{k-1}\frac{x^k}{k}$$ for all $x \in (-1,1]$.
My reference then continues by estimating the error for partial sums of this series:
$$f(x)= \sum_{k=0}^{n-1}\left[(-1)^{k-1}\frac{x^k}{k}\right] + E_{n}(x)$$
Using Taylor's Theorem with remainder in Lagrange's form gives
$$E_{n}(x)= \frac{1}{n} \left\lvert\frac{x}{1+\xi_{n}} \right\rvert ^n$$,
where $\xi_{n}$ is some real number between $x$ and $0$.
Certainly $| E_{n} | \to 0$ as $n \to \infty$ if $|\frac{x}{1+\xi_{n}} | ^n <1.$
We also have $| x | \leq 1$ and $| \xi_{n} |<1$ as the R.O.C. is $R=1$ by the ratio test.
My reference now claims, and this is what I don't understand, that this is true just when $x \in [-\frac{1}{2},1]$.
But surely if, say, $x=-3/4 \notin [-\frac{1}{2},1]$ it could be that and $\xi_{n}=-1/2$ and the inequality is still satisfied. So what am I missing?
Any help is greatly appreciated!
If $x=-3/4$ and $\xi=-1/2$, then $$\left\lvert \frac{x}{1+\xi}\right\rvert=\left\lvert\frac{3/4}{1-1/2}\right\rvert=\frac32$$ so you cannot say $E_n\to 0$.
In fact this doesn't happen, but you need to do more work to show it.