Why does the domain of convergence change after this integration? So, while trying to derive the Taylor expansion for $\ln(x+1)$ around $x=0$, I started by using the well-known power series:
$$\frac{1}{1-x}=1+x+x^2 +x^3 +... \qquad \text{when } \lvert x \rvert < 1$$
Letting $x \mapsto-x$, we get:
$$\frac{1}{1+x}=1-x+x^2 -x^3 + ... \qquad \text{when } \lvert x \rvert < 1$$
So then perfoming this integration gives us:
$$\begin{align*}
 \ln(1+x) = \int_0^x \frac{dt}{1+t} &=\int_0^x (1-t+t^2-t^3+...)dt \\
      &= x-\frac{x^2}{2}+\frac{x^3}{3}- \frac{x^4}{4} +... \qquad \text{when } \vert x \rvert<1
\end{align*}$$
However, the real domain of convergence for this is $-1 < x \leq 1$ as $\ln(2) = 1- \frac{1}{2} +\frac{1}{3} - \frac{1}{4}+ ...$
So my question is why did the domain over which the original function converges in change after this integration and is there a way to predict when this happens given some power series, or should one always check the endpoints again?
 A: Writing
$$\tag{*}\ln(1+x) = \int_0^x \frac{dt}{1+t} =\int_0^x (1-t+t^2-t^3+...)\,dt \\ =  x-\frac{x^2}{2}+\frac{x^3}{3}- \frac{x^4}{4} + \ldots $$
is true for $x < 1$ because the power series converges uniformly on the compact interval $[0,x]$ (within the interval $(-1,1)$ bounded by the radius of convergence) and termwise integration is permissible.    
The series $1-t+t^2-t^3+...$ is neither pointwise convergent at $t=1$ nor uniformly convergent on $[0,1)$.  
As much as one might like to simply substitute $1$ for $x$ in (*) and conclude that
$$\tag{**} \ln 2 =  1-\frac{1}{2}+\frac{1}{3}- \frac{1}{4} + \ldots ,$$
it cannot be justifed without further analysis, i.e. there are other examples of this type where it would be false.
In this case, however,  we can establish that the series on the RHS of (**) is convergent by the alternating series test and by Abel's theorem it follows that 
$$\ln 2 =  \lim_{x \to 1-} \ln(1+x) = 1-\frac{1}{2}+\frac{1}{3}- \frac{1}{4} + \ldots $$
Again -- knowing only the radius of convergence is not enough information to establish this result. 
A: The radius of convergence means that it converges when being strictly below. It doesn't converges when it's strictly higher. It's unknown at that radius. Here the radius $R=1$ is conserved, even if $ln(1+1)$ works.
