Why does the taylor series of $\ln (1 + x)$ only approximate it for $-1I'm looking for an intuitive understanding instead of a formal proof. Thanks for the help.
 A: Intuitively, you can think of the interval of convergence of the Taylor series of $f$ centred at $a$ as the largest symmetric interval centred at $a$ for which the resulting infinite series converges to the function $f$ - here, when I say symmetric, I am disregarding the endpoints as it is possible to have intervals of convergence of the form $[a - R, a + R)$ and $(a - R, a+ R]$. 
Given the interval you are asking about, the Taylor series of $\ln(1+x)$ has been taken at the origin, so the corresponding interval of convergence, disregarding endpoints is of the form $(-R, R)$. As $\ln(1 + x)$ is only defined for $x \in (-1, \infty)$, we have $(-R, R) \subset (-1, \infty)$. The largest the interval $(-R, R)$ can be whilst satisfying this condition is $(-1, 1)$ - in fact, this is what the interval turns out to be. 
However, we must check the endpoints. Well, as $-1 \notin (-1, \infty)$, it can't be in the interval of convergence, and by calculating the Taylor series, you can see that the infinite sum converges for $x = 1$ (it is the alternating harmonic series). So the interval of convergence for $f(x) = \ln(1+x)$ is $(-1, 1]$, that is, the Taylor series converges to $f(x)$ for $|x| < 1$ and $x = 1$.
A: Write $$\log(1+x)= \sum_{n=1}^ \infty (-1)^{n+1}\frac {x^{n}}{n}.$$
Now you can apply the ratio test to get the radius of convergence. Finally, you need to check convergence at the boundary points.
