So, the summability of each $(p_{i,k})_{k \in S_i}$ stems from the hypothesis that $(p_t)_{t \in \overline{S}}$ is summable.
We want to show that the family $\left(\sum_{k \in S_i}{p_{i,k}}\right)_i$ is multipliable.
Let $\mathfrak{m}$ be a monomial. Let $N$ be the (finite) set of monomials lower or equal than $\mathfrak{m}$. There is some finite set $F \subset \overline{S}$ such that for all $t \in \overline{S} \backslash F$, the coefficients of the monomials of $N$ in $p_t$ are zero. Let $J \subset I$ be the (finite) set of all the $i \in I$ such that there is $k \in S_i \backslash \{0\}$ such that $(i,k) \in F$.
Let $c_{\mathfrak{n}}(q)$ be the coefficient of a monomial $\mathfrak{n}$ in a polynomial $q$.
Then, for any finite $J' \supset J$, the $\mathfrak{m}$-coefficient of $\prod_{j \in J'}{\sum_{k \in S_j}{p_{j,k}}}$ is the sum of $$\prod_{j \in J'}{c_{\mathfrak{n}_j}(p_{j,k_j})}$$ over $k \in \prod_{j \in J'}{S_j}$ and monomials $\mathfrak{n}_j$ satisfying $\sum_{j\in J'} \mathfrak{n}_j = \mathfrak{m}$ (we write monomials additively, i.e., we say "sum" for what would usually be called "product"). By the assumption on $J$, $c_{\mathfrak{n}_j}(p_{j,k})$ (the coefficient for
the monomial $\mathfrak{n}_j$ of $p_{j,k}$) is zero when $j \notin J$ and $k \neq 0$, and $1$ if $k=0, \mathfrak{n}_j=0$. So
$$c_{\mathfrak{m}}\left(\prod_{j \in J'}{\sum_{k \in S_j}{p_{j,k}}}\right)=\sum{\prod_{j \in J}{c_{\mathfrak{n}_j}(p_{j,k_j})}},$$ where the sum is over the decompositions $\mathfrak{m} = \sum_{j \in J}{\mathfrak{n}_j}$ and the $k \in \prod_{j \in J}{S_j}$. Thus this coefficient does not depend on the choice of $J' \supset J$.
Therefore the family $\left(\sum_{k \in S_i}{p_{i,k}}\right)_i$ is multipliable.
Let's now show that the family $\left(\prod_{i \in I}{p_{i,k_i}}\right)_k$ is summable, over the essentially finite $k \in \prod_{i \in I}{S_i}$.
Let $\mathfrak{m}$ be a monomial. Let $k \in \prod_{i \in I}{S_i}$ be essentially finite. If $\prod_{i \in I}{p_{i,k_i}}$ has a nonzero $\mathfrak{m}$ coefficient, then there is a decomposition $\sum_{i \in I}{\mathfrak{n}_i}=\mathfrak{m}$ such that for each $i \in I$, the $\mathfrak{n}_i$ coefficient of $p_{i,k_i}$ is nonzero.
Let $N$ be, once again, the set of all monomials that are lower or equal than $\mathfrak{m}$. There is a finite subset $T \subset \overline{S}$ such that if $t \in \overline{S}\backslash T$, then $p_t$ has a zero $\mathfrak{n}$ coefficient for each $\mathfrak{n} \in N$.
Now, the set $J$ of $k \in \prod_i{S_i}$ that are essentially finite such that, for each $i \in I$, $k_i = 0$ or $(i,k_i) \in T$ is finite. Indeed, the set $J_1$ of $i$ such that $T$ meets $\{i\} \times S_i$ is finite; for each $i$, $J^i=\{k \in S_i,\, k=0 \text{ or }(i,k)\in T\}$ is finite, and $J$ injects into $\prod_{i \in J_1}{J^i}$ which is finite. Moreover, if $k \notin J$, then $\prod_{i \in I}{p_{i,k_i}}$ has a zero $\mathfrak{m}$-coefficient.
Thus the family $\left(\prod_{i \in I}{p_{i,k_i}}\right)_k$ is summable.
Let $\mathfrak{m}$ be any monomial. To show that $\prod_{i \in I}{\sum_{k \in S_i}{p_{i,k}}}$ and $\sum_k{\prod_{i \in I}{p_{i,k_i}}}$ have the same coefficient for $\mathfrak{m}$, it is thus enough to show the following:
For any finite subsets $J \subset I$ and $K$ of essentially finite elements of $\prod_{i \in I}{S_i}$, there are subsets $J' \supset J, K' \supset K$ such that $\prod_{i \in J'}{\sum_{k \in S_i}{p_{i,k}}}$ and $\sum_{k \in K'}{\prod_{i \in I}{p_{i,k_i}}}$ have the same coefficient for $\mathfrak{m}$.
So let $J,K$ be our finite subsets. Let $J_1$ be the reunion of the $\{i \in I,\,k_i \neq 0\}$ where $k \in K$. Let $K' \supset K$ be the set of all $k \in \prod_{i \in I}{S_i}$ with support in $J'=J \cup J_1 \supset J$.
Then it's elementary (distributivity for finite products of infinite summable sequences) to show that $\prod_{j \in J'}{\sum_{k \in S_j}{p_{j,k}}} = \sum_{k \in K'}{\prod_{i \in I}{p_{i,k_i}}}$, and in particular they have the same coefficient for $\mathfrak{m}$.
Best Answer
Let's see how we find the first sum which's known as telescoping sum and the other sums are almost the same: the idea is to change the index and then cancel most of the terms
$$\sum\limits_{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=\lim_{N\to\infty}\sum\limits_{n=1}^{N} \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=\lim_{N\to\infty}\sum\limits_{n=1}^{N} \frac{1}{\sqrt{n}}-\sum\limits_{n=2}^{N+1}\frac{1}{\sqrt{n}}\\=\lim_{N\to\infty}\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{N+1}}=\frac{1}{\sqrt{1}}$$