Why is the validity range for Maclaurin Series $\ln(1+x)$ : $-1\lt x\le 1$? I only seem to know that $\ln\;(0$ or any negative real  number)  doesnt exist hence $-1\lt x$ but what about $x \le1$? 
Has this got to do with the convergence of the series? If so, why would convergence make it valid but for any $x$ more than $1$ would make it invalid?
Pardon if there's any mistakes in my understanding because this is relatively new topic for me.
 A: We have that
$$\ln(1+x)=\sum_{k=1}^{\infty}(-1)^{k+1}\frac {x^k} k$$
and by root test we obtain
$$\frac{|x|}{\sqrt[k]k} \to |x|$$
then we need $|x|<1$ and for $x=-1$ we obtain a divergent series $\sum -\frac1k$.
A: The series that you are talking about is$$x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots\tag1$$If $\lvert x\rvert>1$, then $\displaystyle\lim_{n\to\infty}\left\lvert(-1)^{n+1}\frac{x^n}n\right\rvert=\infty$, and therefore $(1)$ diverges. If $x=-1$, then the series is minus the harmonic series, and therefore it diverges too. If $x=1$, then it converges, by  Leibniz's test. And if $\lvert x\rvert<1$, then it converges absolutely; just compare it with the series $\sum_{n=1}^\infty\lvert x\rvert^n$.
The fact that this series is the MacLaurin series of $\ln(x+1)$ is not relevant for this analysis.
A: The "intuitive" explanation of why the MacLaurin series diverges for "large" $x$ is that a polynomial will in any case diverge to $\pm \infty$ according to the highest power of its terms, while $ln(1+x)$ goes to infinity "at a very low power".
That's the same situation as with $1/(1+x)$.
A: For $x\le -1$ the function $\ln(1+x)$ is undefined in real numbers. One does not expect an infinite series converge to something which is undefined.
For $x=1$ the beautiful alternating harmonic series $$1-1/2+1/3-1/4+......$$ converges to $\ln 2$
For $x >1$ the divergence test implies the divergence of the series, so the interval of convergence is $(-1,1]$ 
