Xrfgtjtk

$R$ 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 $$limsup_{nrightarrowinfty}sqrt[n]{|a_n|}leq limsup_{nrightarrowinfty}|frac{a_{n+1}}{a_n}|$$ If I would take a sequence $(a_n)_{ninmathbb{N}}$ for which $$limsup_{nrightarrowinfty}sqrt[n]{|a_n|} < 1 <limsup_{nrightarrowinfty}|frac{a_{n+1}}{a_n}|$$ wouldn't there be a contradiction for the respective power series. Because $$limsup_{nrightarrowinfty}sqrt[n]{|a_n|} < limsup_{nrightarrowinfty}|frac{a_{n+1}}{a_n}|$$ 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) $frac{1}{R}=limsup_{nrightarrowinfty}sqrt[n]{|a_n|}=:tneq 0,infty$

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

$$limsup_{nrightarrowinfty}sqrt[n]{|a_nz^n|}=limsup_{nrightarrowinfty}sqrt[n]{|a_n}||z|$$

Converges absolutely if

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

Diverges if

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

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

$(*) Rightarrow frac{R+frac{1}{t}}{2}$, 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 zin mathbb{C}: |z|<R, sum_{n=0}^{infty}a_nz^n $ converges absolutely.

ii) $frac{1}{R}=limsup_{nrightarrowinfty}|frac{a_{n+1}}{a_n}|=:tneq 0,infty$

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

$$limsup_{nrightarrowinfty}|frac{a_{n+1}z^{n+1}}{a_nz^n}|iff limsup_{nrightarrowinfty}|frac{a_{n+1}}{a_n}||z| $$

Converges absolutely if

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

Diverges if

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

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

$(***) Rightarrow frac{R+frac{1}{t}}{2}$, 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 zin mathbb{C}: |z|<R, sum_{n=0}^{infty}a_nz^n $ converges absolutely.

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

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

Your reading of the quotient criterion is partly wrong. The rule for divergence is that if
$$
liminf_{ntoinfty}frac{|a_{n+1}z^{n+1}|}{|a_nz^n|}=|z|liminf_{ntoinfty}frac{|a_{n+1}|}{|a_n|}>1
$$
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.