TITLE: Distribution with singularities.
QUESTION [5 upvotes]: I need some help to prove that $f$ defined by  $\langle f,\psi\rangle:= \sum_{n=0} ^\infty
 \psi^{(n)}(n)$ is a distribution which has singularities of infinite
 order. Here $\psi$ is a test function that belongs to $ \mathcal D(\Bbb R)$.
Thanks.

REPLY [2 votes]: I will give you the main idea. Then I believe you can deal with the technical details.
You may already suspect that 
$$
f=\sum_{n=0}^{\infty} (-1)^n \delta^{(n)}_n = \mathcal D - \lim_{N\rightarrow \infty}\sum_{n=0}^{N} (-1)^n \delta^{(n)}_n
$$
where the equality and the limit are taken in the sense of distribution. You can verifie that the distribution $S_N = \sum_{n=0}^{N} (-1)^n \delta^{(n)}_n$ is well defined and also its limit $f$. 
Then, since $\delta^{(n)}_x$ is a distribution of order $n$, you have automatically the wanted result.