$R$ 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.
Best Answer
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.