First, it is useful to know that a particular recipe for a function may only be defined on part of that function's domain. For example,
$$ f_1(x) = \frac{1}{1-x} $$
has domain $\Bbb{R} \smallsetminus \{-1\}$. However, a different recipe
$$ f_2(x) = 1 + x + x^2 + x^3 + \cdots $$
has finite values only on $(-1,1)$. (You should recognize that $f_2$ is a geometric series, so you can know the value of the infinite sum.) It agrees with the recipe $f_1$ at each point in that interval, but doesn't even have a value off that interval. (In the jargon: the series doesn't converge for $x \leq -1$ or for $x \geq 1$.) However, there can be more recipes giving (pieces) of the function $f_1$. For instance,
$$ f_2 = \frac{-1}{2} \sum_{n=0}^\infty \left( \frac{-1}{2}\right)^n (x-3)^n $$
(which is another geometric series) only taking finite values on $(1,5)$.
Notice that both intervals, the one for $f_2$ and the one for $f_3$, stop at $1$ and do not include $1$. It's as if the series is telling us "something bad happens at $1$". If we go back to $f_1$ and look at $x = 1$, we find division by zero -- a famously undefined operation. So perhaps we shouldn't expect "nice" recipes like Maclaurin or Taylor series to "work" at "bad" places like $x = 1$. (I've quoted imprecise terms that I'll improve a little, but not all the way to the rigorous end, below.)
Notice that the interval for $f_2$ is symmetric around $0$ and we have $x = x-0$ as the term containing $x$ with successive powers in the sum. The interval for $f_3$ is symmetric around $3$ and we have $x-3$ as the term containing $x$ with successive powers in the sum. This is a generic property of Taylor/Maclaurin sums -- the region of convergence is centered on the offset applied to $x$ in the power term.
These series can be interpreted as taking real $x$s or complex $x$s. For complex $x$s, the disk of convergence (in the complex plane) is centered as described above. The radius is the largest radius that does not include any "bad" points. For $f_1$, the division by zero at $x = 1$ is a bad point (and is the only such point), so no matter where you center your disk, the radius is the distance from your center to $x=1$. Then the disk extends right up to the bad point and no further. (Some interesting things can happen on the boundary of the disk. A series may converge on none, some, or all but one point of the boundary. The set of points of convergence can be complicated and messy. Let's put that off for a few years.) If the center is real, the radius is still controlled by the location of bad points on the complex plane. So we can "not see" the bad points off the real line, but they still impact the interval of convergence. Again, the two endpoints of an interval of convergence may neither, one, or both converge. (For the $f$s, the endpoint at $1$ can never converge because the bad point is on the real line. If the bad point were off the real line, maybe the two endpoints can converge.)
So after that crash course on phenomenology of power series, what about your series?
$$ \ln(1+x) \ \text{"$=$''} \ -\sum_{n=1}^\infty (-1)^n \frac{1}{n} x^n$$
(where the equality symbol is in quotes because we expect generically that the two recipes don't agree everywhere -- the sum may not even give a value everywhere.) We know that $\ln(0)$ is undefined, so there is a bad point at $1+x = 0$, so at $x = -1$. No Maclaurin/Taylor series can converge there. Also, the term containing $x$ with successive powers in the sum is "$x$" (${}=x-0$) so the center is at $x = 0$. From this, we know that largest radius of convergence is $1$ (the distance from $x = 0$ to $x = -1$). Without a course discussing the complex logarithm, I'll reveal that the logarithm has no more bad points (except the one at infinity, which we need not worry about). So the radius of convergence is $1$ and the interval of convergence is either $(-1,1)$ or $(-1,1]$.
A Calculus student studying series would study the $1$ end of the interval by studying the series
$$ \left. -\sum_{n=1}^\infty (-1)^n \frac{1}{n} x^n \right|_{x = 1} = -\sum_{n=1}^\infty (-1)^n \frac{1}{n} \text{,} $$
recognize that the alternating series test applies here, and determine that the series converges at $x = 1$, so the interval of convergence is $(-1,1]$. That is,
$$ \ln(1+x) = -\sum_{n=1}^\infty (-1)^n \frac{1}{n} x^n, x \in (-1,1] \text{.} $$
There is a third question: just because a series recipe converges does not mean that the series converges to the same thing as the function. (Classical example: convergence of a Fourier series at a jump discontinuity of a function.) Questions of whether the values of a series agree with another series or another function are ... more complicated. Most Calculus students aren't exposed to this problem, so aren't aware that a Maclaurin/Taylor series can converge on an interval to something other than the function they're expecting. (To be fair, for the tame-ish functions that Calculus students work with, the series agree just fine. It takes a somewhat wild function to cause problems. The functions here are not wild enough to cause a problem.)
To summarize, the recipe $\ln(1+x)$ is defined on $(-1,\infty)$ and the recipe $-\sum_{n=1}^\infty (-1)^n \frac{1}{n} x^n$ only provides values for $x \in (-1,1]$. So these two recipes can only agree on $(-1,1]$. In fact they do. Nevertheless, the series can only give values on that smaller interval.
