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

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

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

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