Why is domain of convergence of Taylor series of $\ln(x)$ about $x=1$ is $ (0,2)$? I can understand the lower bound as $\ln(x)$ doesn't exist for $x<0$. But how is the upper bound $2$?
 A: Hints:
$$\log x=(x-1)-\frac{(x-1)^2}{2}+\ldots=\sum_{k=1}^\infty\frac{(-1)^{k-1}(x-1)^k}{k}$$
$$\left|\;\frac{(x-1)^{k+1}}{k+1}\;\cdot\frac{k}{(x-1)^k}\right|=|x-1|\frac k{k+1}\xrightarrow [k\to\infty]{}|x-1|$$
It then has to be $\,\;|x-1|<1\iff \,$ ...
Further hint: for  $\;a,b\in\Bbb R\;,\;a>0\;;\;|x-b|<a\iff -a<x-b<a\;\ldots\;$
A: Here's a rule of thumb. A "nice" function $f(x)$, such as a rational function or the logarithm function, can be expanded in a Taylor series about a nonsingular point $x_0$. The radius of convergence at that point is the smallest distance between $x_0$ and a point where $f(x)$ has a singularity (the singularity may be a complex number.) By "singularity" I mean a point where the function shoots off to infinity, usually because of division by zero. For example,


*

*the Taylor series for $1/(1-x)$ about $x=0$ has radius of convergence $1$ because $x=1$ is a singular value.

*the Taylor series for $\log x$ about $x=1$ has radius of convergence $1$ because $x=0$ is a singular value.

*the Taylor series for $1/(x^2+1)$ about $x=0$ has radius of convergence $1$ because $x=i$ is a singular value.


This rule of thumb says nothing about behavior on the boundary of the convergence region (in your example, the points $x=0$ and $x=2$). And again, it is only a rule of thumb.
A: There is the following theorems (in complex analysis):
Given a power series in complex number $z$ about $z_{0}$. Then exactly 1 of the 3 possibilities hold:


*

*The series converge at $z=z_{0}$ and diverge everywhere else.

*The series converge for all $z$.

*There exist an $R\in\mathbb{R}^{+}$ such that for all $z$ where $|z-z_{0}|<R$ then the series converge; and for all $|z-z_{0}|>R$ then the series diverge. As for $|z-z_{0}|=R$ the series could converge or diverge, but we know that there exist at least 1 such $z$ such that the function defined by the series have no Taylor series expansion about $z$.
Another theorem for real number variable only (Abel's theorem):
If a power series $\sum\limits_{n=0}^{\infty}a_{n}x^{n}$ satisfy $\sum\limits_{n=0}^{\infty}a_{n}$ either converge or diverge to infinity, then $\lim\limits_{x\rightarrow 1^{-}}(\sum\limits_{n=0}^{\infty}a_{n}x^{n})=\sum\limits_{n=0}^{\infty}a_{n}$.
Hence "radius of convergence" is a well-defined concept. And it behave exactly like what its name implied. So once you know that the series for $\ln$ about $1$ diverge at $0$, then $R\leq 1$ so the series cannot possibly converge at $z>2$. If you know that the series converge on $(0,1)$ then it is automatically the case that $R=1$ so the domain of converge can only be $(0,2]$ or $(0,2)$. Since the sum of the coefficient for Taylor series of $\ln$ around $1$ converge, the domain of convergence is actually $(0,2]$ (not just $(0,2)$).
