Finding the support of a distribution Definition. Let
 $u :C^{\infty}_{c}(\Omega) \rightarrow \mathbb{C}$ a linear continuous functional, that is, $u$ is a distribution. We define the support of $u$ as $$\Omega- \{x \in \Omega; \exists \; \mbox{an open}\; V_x \; \mbox{such that}\; \; u|_{V_x}=0\}.$$ Here $u|_{V_x}$ means $$u|_{V_x}(\varphi)=u(\varphi)\quad \forall \varphi \in C_{c}^{\infty}(V_x),$$ where $\Omega \subset \mathbb{R}^{n}$ is open.
How can I find the support of the distribution \begin{equation} \label{1}u(\varphi)=\sum_{n=1}^{\infty}[\varphi (\frac{1}{n})-\varphi(0)-\frac{1}{n}\varphi'(0)] \quad \forall \varphi \in C^{\infty}_{c}(\mathbb{R})? \end{equation}
I would like an answer or a tip on how to solve this question.
I observe this serie is convergent, because using the Taylor Formula $$\varphi(\frac{1}{n})=\varphi(0)+\frac{1}{n}\varphi'(0)+\frac{1}{n^2}\alpha(\frac{1}{n})$$ we conclude $u(\varphi)$ is convergent. 
 A: For which $x \in \mathbb{R}$ are the value of $\varphi(x)$ used?
A: We will prove that $$ \mbox{supp}\, u=\{0,1,\frac{1}{2}, \ldots, \frac{1}{n}, \ldots\}:=S.$$
Fisrtly we notice that $S$ is closed in $\mathbb{R}$, because $\frac{1}{n}\rightarrow 0$. Now, if $\varphi \in C^{\infty}_{c}(\mathbb{R}-S)$ then $$ \langle u, \varphi\rangle=\langle u|_{\mathbb{R}-S}, \varphi \rangle=\sum_{n=1}^{\infty}\left(\varphi(\frac{1}{n})-\varphi(0)-\frac{1}{n}\varphi'(0))\right)=0,$$ because $\mbox{supp}\, \varphi' \subset \mbox{supp}\, \varphi$ and $S \cap \mbox{supp}\, \varphi=\varnothing$. Thus, if $x \in \mbox{supp}\, u$ then $x \in S$, because $\mathbb{R}-S$ is open and $u|_{\mathbb{R}-S}=0$. So $$\mbox{supp}\, u \subset S. $$
Now, let $x=\frac{1}{p} \in S$ and $V$ an open subset of $\mathbb{R}$ with $x \in V$. Define \begin{equation}\Omega=
\begin{cases}
V \cap ]\frac{1}{p+1}, \frac{1}{p+1}[ \quad p >1\\
V \cap ]\frac{1}{2},2[, \quad p=1.
\end{cases}
\end{equation} Thus there exists $\varphi \in C^{\infty}_{c}(\Omega)$, $0 \leq \varphi \leq 1$ and $\varphi(x)=1$. Therefore $\varphi(0)=\varphi(\frac{1}{j})=0$ if $j \neq p$, and we obtain \begin{equation}
\langle u|_{V},\varphi \rangle=\langle u, \varphi \rangle=\sum_{n=1}^{\infty}\left(\varphi(\frac{1}{n})-\varphi(0)-\frac{1}{n}\varphi'(0)\right)=\varphi(\frac{1}{p})=1.
\end{equation}
Therefore $x \in \mbox{supp}\, u$ and we have what we wish for, that is,  $$\{1, \frac{1}{2}, \ldots, \frac{1}{n}, \ldots\}\subset\mbox{supp} u .$$
Besides $0 \in \mbox{supp}\, u$, because $\frac{1}{n}\rightarrow 0$ and $\mbox{supp}\, u$ is closed in $\mathbb{R}$.
