Region of convergence for the series of functions Find out the region of convergence for the following series of function.

$$\sum_{m=1}^\infty x^{\log (m)}$$

Here $x \in \mathbb{R}$. This is a series of function. I was trying to find out the radius of convergence by Root Test and Ratio test. But no suitable solution I am getting.

What will be the region of convergence if we consider $x \in \mathbb{C}$?

Thank you for your help.
$x \in \mathbb{R}$ ??. How do we define $(-2)^{log(2)}$ ? And in my opinion, this is not really a power-series .
 As what I solved, this series converges if and only if $ x \in [0; \frac{1}{e})$
Hint: 
Rewrite the series as :
 $ \sum_{k=0}^{\infty}  \sum_{m=[e^k]}^{[e^{k+1}]-1} x^{log(m)} $ (where $[.]$ is floor function) 
And we see that:
$( [e^{k+1}]-1-[e^k]).x^{k-1} \le \sum_{m=[e^k]}^{[e^{k+1}]-1} x^{log(m)} \le ( [e^{k+1}]-1-[e^k])x^{k+1}$(*)
Which leads to the interval of convergence.
One idea would be to use the fact that $\:x^{\log(m)}=e^{\log(m)\log(x)}=e^{\log(m)\Large(\normalsize\log(x)/\log(m)\Large)}\underset{m\to\infty}{\sim}e^{\log(m)}.$