3
$\begingroup$

$R$ is the radius of convergence for a powerseries

I will write down my proof but I am not sure whether this is right because I thought $$\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}\leq \limsup_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|$$ If I would take a sequence $(a_n)_{n\in\mathbb{N}}$ for which $$\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|} < 1 <\limsup_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|$$ wouldn't there be a contradiction for the respective power series. Because $$\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|} < \limsup_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|$$ and the statement in my question would imply $\frac{1}{R}<\frac{1}{R}$


Please tell me where I made the mistake in my reasoning of the following proof:

i) $\frac{1}{R}=\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}=:t\neq 0,\infty$

Applying the root criteria for an arbitrary power series $\sum_{n=0}^{\infty}a_nz^n$

$$\limsup_{n\rightarrow\infty}\sqrt[n]{|a_nz^n|}=\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n}||z|$$

Converges absolutely if

$$\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n}||z|<1\iff |z|<\frac{1}{\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n}|}=\frac{1}{t}\tag{*}$$

Diverges if

$$\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n}||z|>1\iff |z|>\frac{1}{\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n}|}=\frac{1}{t}\tag{**}$$

Suppose $\frac{1}{R}>t\iff R<\frac{1}{t}\Rightarrow R<\frac{R+\frac{1}{t}}{2}<\frac{1}{t}$

$(*) \Rightarrow \frac{R+\frac{1}{t}}{2}$, converges absolutely. Contradiction

Because $R:=\sup\{|z|:\sum_{n=0}^{\infty}a_nz^n$, converges$\}$

Suppose $\frac{1}{R}<t$, $(**) \Rightarrow$ The power series $\sum_{n=0}^{\infty}a_nz^n$ diverges for $z=\frac{R+\frac{1}{t}}{2}$ Contradiction

Because $\forall z\in \mathbb{C}: |z|<R, \sum_{n=0}^{\infty}a_nz^n $ converges absolutely.


ii) $\frac{1}{R}=\limsup_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|=:t\neq 0,\infty$

Applying the quotient criteria for an arbitrary power series $\sum_{n=0}^{\infty}a_nz^n$

$$\limsup_{n\rightarrow\infty}|\frac{a_{n+1}z^{n+1}}{a_nz^n}|\iff \limsup_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}||z| $$

Converges absolutely if

$$\limsup_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}||z|<1\iff|z|<\frac{1}{\limsup_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|}= \frac{1}{t}\tag{***}$$

Diverges if

$$\limsup_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}||z|>1\iff|z|>\frac{1}{\limsup_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|}= \frac{1}{t}\tag{****}$$

Suppose $\frac{1}{R}>t\iff R<\frac{1}{t}\Rightarrow R<\frac{R+\frac{1}{t}}{2}<\frac{1}{t}$

$(***) \Rightarrow \frac{R+\frac{1}{t}}{2}$, converges absolutely. Contradiction

Because $R:=\sup\{|z|:\sum_{n=0}^{\infty}a_nz^n$, converges$\}$

Suppose $\frac{1}{R}<t$, $(****) \Rightarrow$ The powerseries $\sum_{n=0}^{\infty}a_nz^n$ diverges for $z=\frac{R+\frac{1}{t}}{2}$ Contradiction

Because $\forall z\in \mathbb{C}: |z|<R, \sum_{n=0}^{\infty}a_nz^n $ converges absolutely.


i) + ii) $\Rightarrow \limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}=\frac{1}{R}=\limsup_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|$

I am thinking about this for some time now please help me to solve the problem.

$\endgroup$
5
  • $\begingroup$ You probably want to remove the $n$'th power of $a_n$ in each one of its twelve occurrences. $\endgroup$ Commented Dec 29, 2018 at 17:22
  • $\begingroup$ This should help: en.wikipedia.org/wiki/… $\endgroup$
    – Ben W
    Commented Dec 29, 2018 at 17:22
  • $\begingroup$ Yes I will remove them $\endgroup$
    – RM777
    Commented Dec 29, 2018 at 17:22
  • $\begingroup$ What's the radius of convergence of $1+2x+x^2+2x^3+x^4+2x^5+\cdots$? $\endgroup$ Commented Dec 29, 2018 at 17:27
  • 1
    $\begingroup$ See en.wikipedia.org/wiki/Ratio_test and look for inconclusive. $\endgroup$
    – egreg
    Commented Dec 29, 2018 at 18:17

1 Answer 1

3
$\begingroup$

Your reading of the quotient criterion is partly wrong. The rule for divergence is that if $$ \liminf_{n\to\infty}\frac{|a_{n+1}z^{n+1}|}{|a_nz^n|}=|z|\liminf_{n\to\infty}\frac{|a_{n+1}|}{|a_n|}>1 $$ then the power series diverges. This gives you a third radius to consider. Thus you get $$ R_{quot, sup}\le R_{root, sup}\le R_{quot,inf} $$ and only the radius $R_{root, sup}$ of the root criterion gives the exact region of convergence of the power series. If the coefficient quotients to not have a strict limit, the quotient criterion leaves out an annulus where no claim towards convergence or divergence is possible.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .