It can be shown easily that the taylor series of
$\ln(1+x)$ about $x=0$ is,$$\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}$$ and it converges for $-1<x\leq 1$.
However to check if $\ln(1+x)$ is analytic it has to be shown that the remainder term of the taylor series $$R_n(x)= \frac{f^{n+1}(c) x^{n+1}}{(n+1)!}$$, where $c$ is a point between $0$ and $x$ tends to zero.
It can be shown easily that for $0\leq x\leq 1$ the remainder term tends to zero thus the taylor series converges to $\ln(1+x)$
However I struggle to show that the remainder term tends to zero for $-1<x<0$. Indeed for any $-1<-a<0$ let $x\in [a,0)$ be chosen, then $$|f^{n+1}(c)|=\left|\frac{(-1)^{n+2}n!}{(1+c^n)}\right|\leq\frac{n!}{(1-a)^n}$$, Therefore,
$$|R_n(x)| \leq\frac{n!}{(1-a)^n(n+1)!}=\frac{1}{(n+1)(1-a)^n}$$
However the limit of the last expression as $n\rightarrow\infty$ is $\infty$ and not zero.
Can anyone help me show analyticity for $-1<x<0$? Any help is appreciated, Thanks!
Method 1: See Section 3 here.
Method 2: Learn complex analysis. That will make your task trivial, once you have experience with complex power series, since a power series that converges at every point inside an open disc in $\mathbf C$ is analytic at every point there. More generally, the easiest way to prove many properties about real power series is to view them on an open disc in the complex plane so you can take advantage of complex analysis, e.g., a complex-valued function on an open connected set $\Omega$ in the complex plane that has one derivative automatically is infinitely differentiable and even has local power series expansions at each point in $\Omega$.