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. 
 A: $x \in \mathbb{R}$ ??. 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.
A: Let $r=|x|$. Since $\log m\to+\infty$, for every definition of the complex power, $|x^{\log m}|=r^{\log m}\to+\infty$ if $r\gt1$, thus the series diverges. If $r\lt1$ then $r^{\log m}\to0$ and $r^{n+1}\leqslant r^{\log m}\leqslant r^n$ for every $n\geqslant0$ and every $m$ such that $n\leqslant\log m\leqslant n+1$. The number of indices $m$ corresponding to a given $n$ is of the order of $\mathrm e^n$ such hence the positive series $\sum\limits_mr^{\log m}$ converges if and only if the series $\sum\limits_n\mathrm e^nr^n$ converges, that is, when $r\lt1/\mathrm e$. 
Thus, the series $\sum\limits_mx^{\log m}$ diverges when $|x|\geqslant1$ and converges absolutely when $|x|\lt1/\mathrm e$. It diverges when $x$ is a positive real and $x\geqslant1/\mathrm e$. It remains to see whether the series $\sum\limits_mr^{\log m}\mathrm e^{\mathrm i\theta\log m}$ converges (simply) when $1/\mathrm e\leqslant r\lt1$ and $0\lt\theta\lt2\pi$.
A: One idea would be to use the fact that $\:x^{\log(m)}=e^{\log(m)\log(x)}=e^{\log(m)\Large(\normalsize\log(x)/\log(m)\Large)}\underset{m\to\infty}{\sim}e^{\log(m)}.$
