Let $H$ be a subring of $k[[t]]$ which contains all formal sums of its elements. Let $W(H)=\{i_0,i_1,i_2,\ldots\}$ be the semigroup of orders of elements in $H$, where we have $0=i_0<i_1<i_2<\cdots$. Show that for $\underline{\text{any}}$ choice of elements $S_{i_0},S_{i_1},S_{i_2},\ldots$ of elements of $H$ with $\operatorname{ord} S_{i_\ell}=i_\ell$, we have $$H=\{\sum_{\ell=0}^\infty\alpha_\ell S_{i_\ell}\mid\alpha_\ell\in k\}.$$
I saw this question in an algebraic geometry book. I tried to solve this. But I did trivial thing, so I don't write what I did here. This is just self-studying. I want to learn how to solve. Please help me to solve the question. Thank you for helping.
I underlined the sentence. How to assume this $H$ is equal to that form??
I'm assuming that the assumption that $H$ is closed under formal sums means that if you take a sequence $h_0, h_1, h_2, \dots \in H$ such that $\text{ord}(h_0) < \text{ord}(h_1) < \dots$, then $\sum_l h_l$, which is a well-defined element of $k[[t]]$, is in fact an element of $H$.
Nevertheless, as stated, the claim is false. Consider $H = {\mathbb Z}[[t]]$. This is a subring of $k[[t]]$ that is closed under taking formal sums. We have $W(H) = \mathbb N$. Now take $S_0 = 1$, $S_1 = t$, $\dots$. Then $$\{ \sum_l \alpha_l S_l \mid \alpha_0, \alpha_1, \dots \in k \} = k[[t]] \neq {\mathbb Z}[[t]].$$
So, as already suggested by Jyrki Lahtonen, $H$ is probably supposed to be a $k$-subalgebra of $k[[t]]$. Under this assumption, it is trivial that $$\{ \sum_l \alpha_l S_l \mid \alpha_0, \alpha_1, \dots \in k \} \subseteq H.$$ For the other direction, given an element $h$ of $H$, inductively construct a sequence $\alpha_0, \alpha_1, \dots \in k$ such that for every $n \in \mathbb N$, $h$ agrees with $\sum_{i \leq n} \alpha_l S_{i_l}$ upto, at least, degree $i_n$, i.e., such that $h - \sum_{i \leq n} \alpha_l S_{i_l}$ has order greater than $i_n$. (Determine for yourself how you should pick $\alpha_{n+1}$ once you have already constructed $\alpha_0, \dots, \alpha_n$.) Then $h = \sum_k \alpha_l S_{i_l}$, showing the other inclusion.
(Note that this is just a rephrasing of the argument by Jyrki Lahtonen in the comments.)