Why is the validity range for Maclaurin Series $\ln(1+x)$ : $-1\lt x\le 1$?

Question

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.

Answer 1

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.

Answer 2

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 $\sum -\frac1k$.

Answer 3

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 $(-1,1]$

Answer 4

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 $1/(1+x)$.