Taylor expansion of ln(1+x)

Okay, so the question is regarding the relation between the interval of convergence and the interval for which $\displaystyle f(x) = \ln (1+x) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}x^{n}$. I know they are the same interval, but I am trying to show this by showing that the remainder function, $\displaystyle R_{n}$ tends to $\displaystyle 0$ only when $\displaystyle x\in (-1,1]$. So I have to show that $\displaystyle \frac{f^{(n)}(y)}{n!}x^{n}$ tends to $\displaystyle 0$ where $\displaystyle y$ lies between $\displaystyle 0$ and $\displaystyle x$ (if this approach is wrong, please correct me). It can be observed that $\displaystyle f^{(n)}(x)=(-1)^{n+1}(n-1)!(1+x)^{-n}$ so plugging that in I have to show that $\displaystyle \frac{1}{n}|\frac{x}{1+y}|^{n}$ tends to $\displaystyle 0$ (because if $\displaystyle |R_{n}|$ tends to $\displaystyle 0$, then so does $\displaystyle R_{n}$). So the condition for that to tend to zero is that $\displaystyle |\frac{x}{1+y}| \le 1$. And now this is where I get stuck showing that $\displaystyle x$ must lie in $\displaystyle (-1,1]$.

A secondary question is why does this still converge on some interval even though on that interval, not all the derivatives$\displaystyle f^{(n)}(x)$ are not bounded by some single constant $\displaystyle C$ on that interval?