$\ln(1+x)$ -- why does its power series converge for $|x| < 1$? Is it a similar reasoning for the convergence of a geometric series, when $|x|<1$?
...at $-1$, $\log(1+x) = \log(0)$, which is undefined.
For $x < -1$, $\log(1+x)$ is negative, which is also undefined.
So, the left hand inequality of $|x|< 1$ makes sense, but what about the right hand inequality? 


*

*Why do we need $x$ strictly less than 1 for convergence of its power series?  


Probably something obvious that has just slipped my mind.
If you don't mind: same question for $\ln(1-x)$.
 A: By definition $\log (1 + x) = \int_0 ^x \frac{1}{(1+t)} dt$. Integrating term by term the Taylor series of $$\frac{1}{1+x} = \sum_{n=1}^{\infty} (-1)^n x^n ,\,\, \color{red} {|x| < 1}$$
you get 
$$\log (1  +x) = \sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n} , |x| < 1$$
Using Leibniz Theorem this series also converges at $x =1$. You may also show that the function $f : (-1,1] \to \mathbb R$ defined by $f(x) = \sum_{n\geq 1} (-1)^{n+1} \frac{x^n}{n}$ coincides with $\log (1 + x)$ at $x = 1$ using $$\log (1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} + \ldots + (-1)^{n-1}\frac{x^{n-1}}{n} + r_n (x)$$
where $$r_n (x) = (-1)^n\int_0^x \frac{t^n}{1 + t} dt $$
and noticing that for $x =1$ we have $$|r(1)| \leq \int_0^1 t^n dt = \frac{1}{n+1}$$ then $$\lim_{n \to \infty} r_n (1) = 0$$
A: $\begin{align}\ln(1+x)&=\int\frac{1}{1+x}dx=\int\sum_{n=0}^{\infty}{(-1)^n}{x^n}dx\\&=\sum_{n=0}^{\infty}((-1)^n\int x^ndx)\\&=\sum_{n=0}^{\infty}(-1)^n\frac{x^{n+1}}{n+1}+C\\&=(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)+C\end{align}$
Evaluating at $x=0$ gives $0=\ln(1)=C$, so we get $ln(1+x)=(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)$ if $|x|<1$

 
If you use Ratio Test, we'll obtain:
$$\begin{align*}
L=\lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|} &= \lim_{n\to\infty}\frac{\quad\frac{|x|^{n+2}}{n+2}\quad}{\frac{|x|^{n+1}}{n+1}}\\
&= \lim_{n\to\infty}\frac{(n+1)|x|^{n+2}}{(n+2)|x|^{n+1}}\\
&= |x|\lim_{n\to\infty}\frac{n+1}{n+2}\\
&= |x|.
\end{align*}$$

The ratio test states that:



*

*if L < 1 then the series converges absolutely;

*if L > 1 then the series does not converge;

By the Ratio Test, the series converges absolutely if $|x|\lt 1$ and diverges if $|x|\gt 1$.

A: Assuming that you know the Taylor series expansion of $\ln(1+x)$ at $0$ is
$$
\ln(1+x)=\sum_{n\ge1}(-1)^{n+1}\frac{x^n}{n},
$$
the ratio test for $x\ne0$ can be applied: since
$$
\left|\frac{(-1)^{n+2}x^{n+2}/(n+1)}{(-1)^{n+1}x^n/n}\right|=
|x|\frac{n}{n+1}
$$
and
$$
\lim_{n\to\infty}|x|\frac{n}{n+1}=|x|
$$
we know that the series converges for $|x|<1$ and doesn't converge for $|x|>1$.
It obviously doesn't converge for $x=-1$, because we get the (opposite of the) harmonic series; for $x=1$, the Leibniz test applies and the series converges. It can be proved that, for $x=1$, the series indeed converges to $\ln2$.
As to “why” the series only converges in this interval, the answer can be found in the theory of analytic functions.
The function $\ln(1+x)$ can be extended to an analytic function in the complex plane provided we exclude the set of complex numbers $z$ with $\Re(z)\le-1$ and $\Im(z)=0$. Such an analytic function has a singularity at $-1$, hence the Taylor expansion around $0$ cannot have a radius of convergence greater than $1$ (and actually has $1$ as radius of convergence).
