# 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 $2$?

• Do you know what the series is? Check it... Jun 2, 2013 at 10:58
• I'm leaning the series and got struck at this. Jun 2, 2013 at 11:00
• See "radius of convergence" in your textbook. Jun 2, 2013 at 12:39

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 $\;a,b\in\Bbb R\;,\;a>0\;;\;|x-b|<a\iff -a<x-b<a\;\ldots\;$

• So, suppose if we are evaluating the series at x=2, then the domain of convergence will be (1,3) ? Jun 2, 2013 at 11:17
• No, at $x=2$ the radius of convergence is 2, so the real domain of convergence is $(0,4)$, or perhaps $(0,4]$. Jun 2, 2013 at 12:40
• @ganessh: expanding at $x=a$ gives \begin{align}\log(x)&=\log(a)+\log(x/a)\end{align} and the series for $\log(x)$ expanded at $x=a$ becomes, up to a constant, the series for $\log(x/a)$ expanded at $x/a=1$. Therefore, the interval of convergence is $(0,2a]$.
– robjohn
Jun 12, 2014 at 1:46

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.

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:

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

2. The series converge for all $z$.

3. 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 $(0,2)$).

• I don't think 3. is completely accurate. The series $$\mathrm{Li}_2(z)=\sum_{k=1}^\infty\frac{z^k}{k^2}$$ has radius of convergence $1$, yet for no $|z|=1$ does the series diverge. See Polylogarithm.
– robjohn
Jun 12, 2014 at 1:32
• @robjohn: oh right, I just checked my book again, I misinterpreted it (I thought "cannot be analytically continued" to means it diverge). I'm going to fix this. Thanks.
– Gina
Jun 12, 2014 at 2:10