We consider the derivative operator $\mathrm{D}$ on the space of smooth and rapidly decreasing function $\mathcal{S}$. We denote by $P_n = \frac{1}{0!} + \frac{X}{1!} + \frac{X^2}{2!} + \cdots + \frac{X^n}{n!}$ and $\mathrm{T}_n = P_n(\mathrm{D})$.
My question is the following. Can we show that the sequence of operators $(\mathrm{T}_n)$ converges to some operator $\mathrm{T}$ from $\mathcal{S}$ to itself (that we could then call $\mathrm{T}= \exp(\mathrm{D})$)? If so, can we easily extend that operator over $\mathcal{S}'$?
I suspect, if the result is true, that we can prove it showing that the sequence $(\mathrm{T}_n)$ is a Cauchy sequence in the complete topological vector space $\mathcal{L}(\mathcal{S})$ of linear and continuous operator from $\mathcal{S}$ to $\mathcal{S}$. The topology of the space $\mathcal{S}$ is complete for the usual family of semi-norms $$\lVert f \lVert_{n,m} = \sup_{x\in \mathbb{R}} | (1+|r|^m) D^{(n)} f(r) |.$$
Thanks for your attention.