Convergence for log 2 
*

*The series for $\log (1+x)$ is convergent(by estimating radius of convergence) only with $|x| <1$. Then how is it still true for $x = 1$?

*How is it still apparently true when $x$ is a roots of unity other than $-1$? Such as in the answers to this question: Summing up the series $a_{3k}$ where $\log(1-x+x^2) = \sum a_k x^k$
 A: Abel's theorem is the key to why you can say that the limit as you approach the boundary comes out to the same as evaluating the series at the boundary.  Sivaram's answer addresses why the series converges at those boundary points.
Explicitly, Abel's theorem implies that 
$$\lim_{r\nearrow 1}\log(1+zr)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}z^n$$
for each complex number $z$ with $|z|=1$ such that the series converges.  By continuity of the logarithm, this will be $\log(1+z)$ in such cases.
A: Look up alternating series test or generalized alternating series test (Also known as Dirichlet's test as George pointed out).
A: Here is a way of seeing this result directly without using Abel's theorem. We have
$$
\log(1+x)=\int_{1}^{x+1}\frac{1}{t} \, dt=\int_{0}^{x}\frac{1}{t+1} \, dt \, .
$$
Using the fact that for all $n\in\mathbb N$,
$$
\frac{1}{1+t}=1-t+t^2-t^3+\dots+(-1)^n\frac{t^n}{1+t} \, ,
$$
we obtain the following expression for $\log(1+x)$, with the remainder term expressed as an integral:
$$
\int_{0}^{x}\frac{1}{1+t} \, dt=x-\frac{x^2}{2!}+\frac{x^3}{3!}-\frac{x^4}{4!}+\dots+(-1)^{n-1}\frac{x^{n-1}}{(n-1)!}+(-1)^n\int_{0}^{x}\frac{t^n}{1+t} \, dt
$$
Now note that
$$
0\le\int_0^x\frac{t^n}{1+t} \, dt\le\int_{0}^{x}t^{n-1} \, dt=\frac{x^n}{n}
$$
and so in the case $x=1$, the remainder term tends to $0$ as $n\to\infty$. Hence the power series for $\log2$ does indeed converge to $\log 2$.
