Let $S_n = \sum_{i=1}^n i^{\alpha} (\delta_{1/i}-\delta_{-1/i})$. These are distributions of order zero for all $n \in \mathbb{N}$ and for all $\alpha \in \mathbb{R}$. I know that, if $\alpha < 0$ then $S_n$ sequence converges in $D'(\mathbb{R})$ to a distribution $S \in D'(\mathbb{R})$. Moreover, if $\alpha < -1$ then $S$ has exactly order zero (because $\sum_{i=1}^{\infty} i^{\alpha}$ converges).
Now, if $-1 \leq\alpha < 0$ I claim that the order is exactly $1$.
My idea is to find $\phi_{\epsilon} \in C_c^{\infty} (\mathbb{R})$ such that $1 = | \langle S,\phi_{\epsilon} \rangle| \leq C \Vert \phi_{\epsilon}\Vert_{C^0} \leq C(\text{something that goes to zero as $\epsilon$ goes to zero})$. My problem is I can't find such functions; my only idea is using functions such that $\phi_{\epsilon}(x) = x$ for every $\epsilon$ and thus $| \langle S,\phi_{\epsilon} \rangle|$ is constant, but this seems not to work as well as i thought. Of course, I think I should take $\phi_{\epsilon}$ to be odd. Some hint?
There is also the probelem of $\alpha = -1$. For this has value may the distribution has order zero?