$u$ harmonic and $\frac{u(x)}{|x-x_0|^n}$ bounded, then $u$ identically $0$ 
Let $\Omega\subset\mathbb{R}^N$ be a connected open set and $u$ an
  harmonic function in $\Omega$. Suppose that there exists
  $x_0\in\Omega$ such that there is a neighborhood $V\subset\Omega$ of
  $x_0$ such that, for all positive integer $n$, the function
  $$\frac{u(x)}{|x-x_0|^n}$$ is bounded in $V-\{x_0\}$. Show that $u$ is
  identically $0$ in $\Omega$

This looks related to the theorem of removable singularities. In this paper it uses an argument to show that an harmonic function is null.
I tried to do the same. Since $u(x)/|x-x_0|^N$ is bounded in a neighborhood of $x_0$, for some $M$ there exists $\delta>0$ such that $$|x-x_0|<\delta \implies u(x)<M|x-x_0|^N$$
In the paper it'd argue that RHS is harmonic, but in this case it's not, so I couldn't apply the same reasoning.
I'm also trying to think about the maximum principle. I know that this function is bounded and I have a neighborhood, but I don't know its boundary.
UPDATE:
I had an idea:
Since $\frac{u(x)}{|x-x_0|^n}$ is bounded, it means that 
$$|x-x_0|<\delta \implies\left|\frac{u(x)}{|x-x_0|^n}\right|<M\implies |u(x)|\le M|x-x_0|^n$$
since it's bounded for all $n$, we can make $x$ close to $x_0$ making it less than $1$ and take the limit to make $|u(x)|\le 0$. So the function is $0$ in a neighborhood. I guess that the fact that it is harmonic and $\Omega$ is connected, then it must be $0$ everywhere. Could somebody point to me how to finish it and if this reasonin is ok? 
How to fit the harmonic and connected hypothesis to prove it is $0$ in $\Omega$?
 A: We first use the taylor series around $x_{0}$. WLOG $x_{0}=0$. (This expansion exists since $u$ is analytic on the domain of definition)
$u(x)=u(0)+Du(0)x+ r(x)= Du(0)x+r(x)$
where the function $\frac{r(x)}{|x|}\to0 $ as $|x|\to 0$ . I will show that $Du(0)$ has norm equal to zero as a linear operator. Indeed,
$\frac{u(x)}{|x|} = Du(0)(\frac{x}{|x|})+\frac{r(x)}{|x|}$
Let $\{x_{n}\}_{n\in \mathbb{N}}$ be any sequence converging to zero. Then we have 
$\lim_{n\to\infty}\frac{u(x_{n})}{|x_{n}|}=0=\lim_{n\to\infty} Du(0)(\frac{x_{n}}{|x_{n}|})$. So $||Du(0)||=0$, as desired.
For higher order derivatives, use induction and a higher order taylor approximation:
$u(x)=\sum_{|\alpha|=0}^n\frac{1}{\alpha!}\frac{\partial^\alpha u(0)}{\partial x^\alpha}x^\alpha+\sum_{|\alpha|=n+1}R_{\alpha}(x)x^\alpha=\sum_{|\alpha|=n}\frac{1}{\alpha!}\frac{\partial^\alpha u(0)}{\partial x^\alpha}x^\alpha+\sum_{|\alpha|=n+1}R_{\alpha}(x)x^\alpha$
Finally, by analyticity, the set $u^{-1}(0)$ is open and closed. Since the domain is connected, $u$ vanishes identically on the domain.
