Show that $u$ is not in $\mathscr{D}^{\prime}_{\mathscr{F}}(\Omega)$, i.e. not a distribution of finite order Let $\Omega = (0,1)$. Then the linear functional 
\begin{equation}
\langle u, \varphi\rangle=\sum_{j=1}^{\infty} \varphi^{(j)}\left(\frac{1}{j}\right) \quad \text { for every } \quad \varphi \in \mathscr{D}(\Omega)
\end{equation}
is in $\mathscr{D}^{\prime}(\Omega)$ but not in $\mathscr{D}^{\prime}_{\mathscr{F}}(\Omega)$
I have shown that the above linear functional is a distribution, nonetheless, I don't know where to start in order to show that it is not a distribution of finite order, i.e. $\mathscr{D}^{\prime}_{\mathscr{F}}(\Omega)$. 
It follows from the definition of distribution of finite order that for every compact K $\subset$ $\Omega$ there are numbers $C_{K}$, $m_{k}$ such that 
\begin{equation}
|u(\varphi)| \leqq C_{K} \sum_{|\alpha| \leq m_{K}} \sup _{K}\left|\partial_{x}^{\alpha} \varphi\right|, \qquad \text { for every } \varphi \in C_{0}^{\infty}(K)
\end{equation}
To this end let K be a compact set of $\Omega$ such that
$$
\begin{split}
|\langle u, \varphi\rangle| &= \left|\sum_{j=1}^{\infty} \varphi^{(j)}\left(\frac{1}{j}\right) \right| \leq \sum_{j=1}^{\infty} \left|\varphi^{(j)}\left(\frac{1}{j}\right)\right| \\
& = \sum_{j=1}^{\infty} \left|\partial_x^{\alpha}\varphi \left(\frac{1}{j}\right)\right| \leq \sum_{j=1}^{\infty} \sup_{K} \left\|\partial_x^{\alpha}\varphi \left(\frac{1}{j}\right)\right\| = \left|\frac{1}{j}\right|\sum_{j=1}^{\infty} \sup_{K} \|\partial_x^{\alpha}\varphi\|
\end{split}
$$ 
Is the above true? Then as we approach infinity we obtain $0$ for the $C_{k}$ if we regard them as follows $C_{K} = |\frac{1}{j}|$ Then for a large number $\left(\frac{1}{j}\right)$ is close to zero, which means that the sum is outside of the supp $\phi$ then the function and all its possible derivatives vanish. Thus, we have a $C_{k} = 0$. Thus $u$ cannot be in $\mathscr{D}^{\prime}_{\mathscr{F}}(\Omega)$.
 A: A distribution $u\in\mathcal{D}'((0,1))$ is of finite order if it defines a continuous linear form on the space $\mathcal{D}^{m}((0,1))$ of compactly supported $m$-times continuously differentiable functions for some natural number $m$. Here, the crucial part of this condition is continuity. By construction of the topology of $\mathcal{D}^{m}((0,1))$, a linear form $$u\colon \mathcal{D}^{m}((0,1))\to \mathbb{C}$$ is continuous if for every compact subset $K\subset (0,1)$ there is a constant $C_K>0$ such that $$|\langle u,\varphi\rangle| \leq C_K \sup_{|k|\leq m}\sup_{x\in K} |\varphi^{(k)}(x)|$$ for all $\varphi\in\mathcal{D}^{m}((0,1))$.
More precisely, the topology of $\mathcal{D}^{m}((0,1))$ is the topology of the (locally convex) inductive limit of the spaces $\mathcal{D}^{m}_K((0,1))$ of $m$-times continuously differentiable functions whose support is contained in the compact set $K\subset (0,1)$. The space $\mathcal{D}^{m}_K((0,1))$ carries the topology of uniform convergence of all derivatives up to order $m$.
Using the above charaterisation it should be rather easy to show that the distribution you consider is not of finite order.
