The problem is to determine for what range of values of $\delta$ can we write
$$\ln(x+1)=\sum_{k=1}^{n-1} (-1)^{k-1} \frac{x^{k}}{k}+o(x^{\delta})$$ as $x$ approaches zero.
It seems since $|R_{n}(x)|\leq\frac{x^{n}}{n}$ we can use the definition of "$o$" to try and satisfy $$\lim_{x \to 0}\frac{\frac{x^{n}}{n}}{x^{\delta}}=0$$ Which clearly works for all values of $\delta \in (-\infty,n)$ Is that correct? Or is the estimate of the remainder to harsh
Here's how to derive that inequality.
$\sum_{k=0}^{n-1} x^k =\dfrac{1-x^n}{1-x} $, so $\sum_{k=0}^{n-1} (-1)^kx^k =\dfrac{1-(-1)^nx^n}{1+x} =\dfrac{1}{1+x}-\dfrac{(-1)^nx^n}{1+x} $.
Integrating from $0$ to $t$, $\sum_{k=0}^{n-1} \int_0^t(-1)^kx^k dx =\int_0^t\dfrac{1}{1+x}-(-1)^n\int_0^t\dfrac{x^ndx}{1+x} $ or $\sum_{k=0}^{n-1} (-1)^k\dfrac{x^{k+1}}{k+1} =\ln(1+t)-(-1)^n\int_0^t\dfrac{x^ndx}{1+x} $ or $\ln(1+t)-\sum_{k=0}^{n-1} (-1)^k\dfrac{x^{k+1}}{k+1} =(-1)^n\int_0^t\dfrac{x^ndx}{1+x} $.
If $t > 0$ then $\int_0^t\dfrac{x^ndx}{1+x} \lt \int_0^tx^ndx =\dfrac{t^{n+1}}{n+1} $ and $\int_0^t\dfrac{x^ndx}{1+x} \gt \int_0^t\dfrac{x^ndx}{1+t} =\dfrac{t^{n+1}}{(n+1)(1+t)} $.