$\ln\left(1+x\right)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dots$ when $|x|<1$ or $x=1$. $\ln\left(1+x\right)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dots$ when $|x|<1$ or $x=1$. Why is the restriction $|x|<1$ or $x=1$? I know from Wikipedia that it is because out of this restriction, the function's "Taylor approximation" is not fair. But how did we derive this limit?
 A: The written form above can be reduced to this one: $$f(x)=\ln |1+x|=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}$$ When we read the chapter of infinite series in any advanced calculus book, we see a section called The Power series. In this section, we learn how to use these kind of series in Calculus. Meanwhile, we learn if a function $f(x)$ and its derivatives $f'(x),f''(x),\cdots,f^{(n)}(x)$ exist and are continuous in interval $I=[a,b]$ and also $f^{(n+1)}(x)$ exists in interval $(a,b)$; then the function can be written as follows: $$f(x)=f(a)+f'(a)(x-a)+\cdots+\frac{f^{(n)}(x)}{n!}(x-a)^n+R_n$$ in which $R_n$ is the reminder. For a suitable interval $I$ we can check that $R_n\to 0$. Note that in this case we choose $a=0$. Now, what is this proper interval? Indeed, the ratio test can be used here to give us that interval.
A: There are many good answers online about this. A brief description is that any series on the form:
$\sum_{n=0}^{\infty}c_n z^n$ 
for constants $c_n$ will converge at $|z|<r$, possible converge or diverge at $|z|=r$ (on a case-by-case basis), and diverge at $|z|>r$, for some $r$. This holds for complex numbers where $|z|$ is taken to be the norm of the complex number, or real numbers where $|z|$ is taken to be the absolute value.
In a series like $\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!} x^{2n+1}$ we have $r=\infty$, and in a series like $\sum_{n=0}^{\infty} n^n x^n$ we have $r=0$. ($6!=6*5*4*3*2*1$ is the factorial symbol)
Wolfram MathWorld
Wikipedia
A: The standard approach is to consider the derivatives of both functions. On the right you obtain $\sum(-x)^n$, which you may recognize as a geometric series, with sum $\frac1{1+x}$, the formula being valid for $|x|<1$. But this is also the derivative of the function on the left. This means the two functions are equal, up to a constant. Evaluate both functions at $x=0$ to see that the constant is zero.
The above leaves a few blanks, of course. Notably, that the power series converges for $|x|<1$ and that we can differentiate it term by term in this interval. This follows from general results on the radius of convergence of power series, and on uniformly convergent sequences of functions. 
What the argument does not show is that the equality also holds at $x=1$. This needs a separate proof, and follows from a general theorem due to Abel. Since the series diverges when $x=-1$, the equality does not hold (for $x$ real) beyond the region $-1<x\le 1$.
A: I think your question is why $|x|<1$ and x=1. The reason is that the formula is not valid for $x=-1$. So we can't say $|x| \le 1$ as we want to exclude $x = -1$.
The limit has to be less than $1$ as $\log(1+x)$ blows up at $-1$. To show that we can get this close to $1$ requires more analysis.
