Consider the series $\sum _{n=0}^{\infty} (-1)^n z^{2n+1}=z-z^3+z^5-z^7+...$
We notice that $a_k=0$ if $k=2n$ and $a_k=(-1)^n$ if $k=2n+1$, thus $\nexists \lim |a_n|^{\frac{1}{n}}=L$.
However the series has a radius of convergence.
But if I'm not wrong, the radius of convergence is defined as $R=\frac{1}{\lim |a_n|^{\frac{1}{n}}}$. So how can $R$ exist when $L$ does not exist?

2. If $z=1$ then $\sum_{k=0} ^{ \infty} (-1)^k z^{2k+1}$ does not converge, but $\vert \sum_{k=0} ^{ \infty} (-1)^k z^{2k+1} \vert \leq \sum_{k=0} ^{ \infty} \vert (-1)^k z^{2k+1} \vert \leq \sum_{k=0} ^{ \infty} \vert z^{k} \vert$ and this last one converges whenever $\vert z \vert <1$ so the radius of convergence is 1.

3. Originally Posted by Jose27
If $z=1$ then $\sum_{k=0} ^{ \infty} (-1)^k z^{2k+1}$ does not converge, but $\vert \sum_{k=0} ^{ \infty} (-1)^k z^{2k+1} \vert \leq \sum_{k=0} ^{ \infty} \vert (-1)^k z^{2k+1} \vert \leq \sum_{k=0} ^{ \infty} \vert z^{k} \vert$ and this last one converges whenever $\vert z \vert <1$ so the radius of convergence is 1.
I get what you mean, all is fine.
But still, I don't understand why $R$ is defined as $\frac{1}{\lim |a_n|^{\frac{1}{n}}}$, it doesn't make sense. Otherwise $R$ wouldn't exist if $L$ doesn't exist, but this is not the case.

4. The definition of the radius of convergence is: If $s= \limsup {\vert a_n \vert ^{ \frac{1}{n} }}$ then $R= \frac{1}{s}$ and notice that in the series you got $s=1$.