This is the definition of convergence in Schwartz space:
We say that $f_n \rightarrow f$ in $S$ if $\lim_{n \to \infty} ||f-f_n||_{ \alpha , \beta} \rightarrow 0$ for all $ \alpha , \beta \in \mathbb{N_{0}^{d}}$
Where,
The Schwartz space $S( \mathbb{R^d}) $ is the set of functions $f \in C^{\infty}(\mathbb{R^d})$ such that:
$$||f||_{ \alpha , \beta} = || x^{\alpha} \partial_{x}^{\beta} f||_{\infty} < \infty$$
for all multiindices $ \alpha , \beta$.
Proposition: ($S$ is a metric space). Convergence in $S$ is equivalent to convergence with respect to the metric:
$$d_{S}(f,g) = \sum_{n=0}^{\infty} 2^{-n} \sup_{ | \alpha| + | \beta | = n} \frac{ ||f-g||_{ \alpha , \beta}}{1+||f-g||_{\alpha,\beta}}$$
Where,
The Schwartz metric,
Now, maybe I'm confused or probably overthinking, but when I'm thinking 'convergence' in $d_{S}$, this is independent on $n$ (the index of the sum) Please correct me if I'm wrong.
However, suppose $d_{S}(f_{j},f) \rightarrow 0$, since each term of the sum is positive, all of the terms have to approach zero, which means that $||f_j -f||_{\alpha , \beta} \rightarrow 0$ $\forall \alpha , \beta$ since the $sup$ itself approaches $0$
Now the notes do not use this method of proof which is supposed to be trivial, so I'm starting to suspect that there's something wrong with my reasoning.
Your problem has not much to do with the Schwartz space. For an increasing sequence $p_n$ of seminorms on a vector space $X$ you want to show that a sequence $(x_k)_{k\in\mathbb N}$ converges with respect to each $p_n$ to a vector $x$ if and only if it converges with respect to $d(x,y)=\sum\limits_{n=1}^\infty 2^{-n} \frac{p_n(x-y)}{1+p_n(x-y)}$. The crucial point is that the condition $d(x,y)<\varepsilon$ only depends on the first $n(\epsilon)$ terms of the series. It seems that this form of the metric was first used by Fréchet who also introduced another one, namely $$D(f,g)=\sup\{p_n(x-y) \wedge 1/n: n\in\mathbb N\} $$ where $a\wedge b$ denotes the minimum of $a,b\ge 0$. Here it is much easier to see that $D(x,y)<1/n$ if and only if $p_n(x-y)< 1/n$ (to check topological concepts, it is enough to consider $\varepsilon=1/n$).