# Finding the radius of convergence with limsup

For a power series defined as

\begin{aligned}\sum_{n=0}^{+\infty} \end{aligned}a_n\,(z-z_0)^n

the radius of convergence is equal to

$R = \frac{1}{\limsup\limits_{n\rightarrow+\infty} \sqrt[n]{\left|a_n\right|}}$

that, if it exists, is equivalent to

$R = \frac{1}{\lim\limits_{n\rightarrow+\infty} \sqrt[n]{\left|a_n\right|}}$

My question then is the following: you might propose to me a power series where the latter limit does not exist and therefore must refer to the strict definition?

Thank you!

• $$a_{2n}=2^n\qquad a_{2n+1}=1$$ – Did Dec 2 '16 at 20:56
• You could just as well have $a_{2n+1}=0$... – Thomas Andrews Dec 2 '16 at 21:01
• That is not recurrence. It is a sequence that alternates between $1$ and powers of $2$. – Fimpellizieri Dec 2 '16 at 21:10
• @Manu Please read carefully what is posted to answer your question, before commenting on it. – Did Dec 2 '16 at 21:11
• $a_n=(\sin(n))^n$ – hamam_Abdallah Dec 2 '16 at 21:28