# Is $\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}=\frac{1}{R}=\limsup_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|$?

$$R$$ 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 $$\frac{1}{R}, $$(**) \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| 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 $$\frac{1}{R}, $$(****) \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| converges absolutely.

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

• You probably want to remove the $n$'th power of $a_n$ in each one of its twelve occurrences. – uniquesolution Dec 29 '18 at 17:22
• What's the radius of convergence of $1+2x+x^2+2x^3+x^4+2x^5+\cdots$? – Lord Shark the Unknown Dec 29 '18 at 17:27
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.