The expression is not a distribution on $\mathbb{R}$

Question

My problem is

Show that $$u(\varphi)=\sum_{n=0}^{\infty}\frac{1}{n!} \varphi^{(n)}(0)$$ doesn't define a distribution on $\mathbb{R}.$

I don't know where to start. I would like an answer.

thank you.

Answer 1

Assuming $\varphi$ has a Taylor series expansion centered at $0$, which series converges at $1$, you seem to have written $\varphi(1)$ in a complicated way.