Show that $\sum\limits_{n=-\infty}^{\infty}\delta^{(|n|)}(x-n)$ diverges in S'

by Schwartz’s theorem, any generalized function from $$S'$$ has a finite singularity order. In this example, it is infinite and I want to show that the series $$\notin S'$$.
($$g^{(l)}$$ means $$l$$th derivative)

Let $$f=\sum\limits_{n=-\infty}^{\infty}\delta^{(|n|)}(x-n)$$ then for each $$\phi \in S',$$ $$(f,\phi)<\infty$$ and f is linear.

I try to give an example when $$f$$ is not continuous.
(I remind you that $$\phi$$ are rapidly decreasing functions and $$\phi_k \to \phi\:$$ iff $$\:\forall l \: \phi_k^{(l)} \rightrightarrows \phi^{(l)}$$ in $$\mathbb{R}$$).

So I try to take $$\phi = 0$$ (maybe it's easyer), then $$\phi_k \to 0$$ and$$(f,\phi_k)=\sum\limits_{n=-\infty}^{\infty}\phi_k^{(|n|)}(n)$$ I want to $$(f,\phi_k) \nrightarrow 0$$.
It looks like $$\phi_k$$ should look something like $$\frac{n^ae^{-bx}}{k}$$ but it doesn't work...
Please give a hint how to find the right $$\phi_k$$