Convergence in Schwartz Space The seminorms are on $\mathcal{S(\mathbb{R}^k)}$ are given by $p_{\alpha, \beta}(\phi) = \sup \vert x^\alpha \partial^\beta\phi \vert$, and the metric is given by
$$
d(\phi, \psi) = \sum_{\alpha, \beta \in \mathbb{N}^k} \frac{p_{\alpha, \beta}(\phi - \psi)}{ 1 + p_{\alpha, \beta}(\phi - \psi)} 2^{- \vert \alpha \vert - \vert \beta \vert}
$$
I want to show that $p_{\alpha, \beta}(\phi_0 - \phi) \to 0$ for all $\alpha, \beta$  implies $d(\phi_n, \phi) \to 0$ , but I am having a hard time finding a rigorous $\delta-\epsilon$ language.
I can rewrite $d(\cdot, \cdot)$ as
$$
d(\phi, \psi) = \sum_{M = 0}^\infty 2^{-M} \sum_{\vert \alpha  \vert + \vert \beta \vert = M } \frac{p_{\alpha, \beta}(\phi - \psi)}{ 1 + p_{\alpha, \beta}(\phi - \psi)} 
$$
The problem occurs because each term in the sum $\sum_{\vert \alpha  \vert + \vert \beta \vert = M } \frac{p_{\alpha, \beta}(\phi - \psi)}{ 1 + p_{\alpha, \beta}(\phi - \psi)}$ is less than 1, but the whole sum may not, which makes it difficult.
Another fact I am trying to use is that for each fixed $M$, there can be at most $(2k)^M$ choices for $(\alpha, \beta)$ such that $\vert \alpha \vert + \vert \beta \vert = M$
 A: For each $\epsilon>0$, and for each $\alpha,\beta$, if we define
$$ c_{k} = \frac1{\#\{(\alpha,\beta) : |\alpha|+|\beta| \le k \}}$$there is  $N_{k,\alpha,\beta}$ (depending on $\epsilon$) such that if $n>N_{k,\alpha,\beta}$ then $p_{\alpha,\beta}(f_n,f) < c_{k}\cdot \epsilon $. Define
$$ N_k := \sup_{|\alpha+|\beta| \le k} N_{k,\alpha,\beta}.$$ We split $d$ into two parts
$$ d(f,f_n) =  \sum_{|\alpha|+|\beta| \le k} \frac{p_{\alpha,\beta}(f,f_n)}{1+p_{\alpha,\beta}(f,f_n)}2^{-|\alpha|-|\beta|} + \sum_{|\alpha|+|\beta| > k} \frac{p_{\alpha,\beta}(f,f_n)}{1+p_{\alpha,\beta}(f,f_n)}2^{-|\alpha|-|\beta|} =: d_{1k}+d_{2k}$$
$d_{1k}$ is controlled once $n>N_k$:
$$\frac{p_{\alpha,\beta}(f,f_n)}{1+p_{\alpha,\beta}(f,f_n)}2^{-|\alpha|-|\beta|} \le p_{\alpha,\beta}(f,f_n) \le c_{k}\cdot \epsilon  \implies d_{1k} \le \epsilon$$
$d_{2k}$ determines the choice of $k$: since $\frac{p_{\alpha,\beta}(f,f_n)}{1+p_{\alpha,\beta}(f,f_n)}\le 1$, and $\sum_{\alpha,\beta} 2^{-|\alpha|-|\beta|} = C<\infty$ where $C$ only depends on dimension, we have that  $d_{2k}  \to 0$.
We put these together in the following way. Fix $\epsilon>0$. Choose $k$ so that $d_{2k} \le \epsilon$. Then for all $n>N_k$, $d_{1k} < \epsilon$. So $d(f,f_n) < 2\epsilon$, QED.
