[proofplan] We use the definition of $\sum_{i \in I} N_i$ as the collection of finite sums of elements from the submodules $N_i$. Closure under subtraction follows by concatenating one finite expression for the first element with the negatives of a finite expression for the second. Closure under scalar multiplication follows by distributing the scalar over a finite sum and using that each $N_i$ is a submodule. The containment and leastness properties then follow directly from the one-term finite-sum description and from the closure of any containing submodule under finite addition. [/proofplan]

[step:Use finite sums to verify the submodule conditions]Set \begin{align*} S := \sum_{i \in I} N_i. \end{align*} By the empty-sum convention, $0 \in S$, so $S$ is nonempty. Let $x,y \in S$ and let $r \in R$. By the definition of $S$, there exist nonnegative integers $m,n$, indices $i_1,\dots,i_m \in I$ and $j_1,\dots,j_n \in I$, and elements $x_k \in N_{i_k}$ for $1 \le k \le m$ and $y_\ell \in N_{j_\ell}$ for $1 \le \ell \le n$, such that \begin{align*} x = \sum_{k=1}^{m} x_k \end{align*} and \begin{align*} y = \sum_{\ell=1}^{n} y_\ell. \end{align*} If $m=0$ or $n=0$, the corresponding sum is interpreted as $0$. For each $\ell$, since $N_{j_\ell}$ is a submodule of $M$, it is closed under additive inverses, so $-y_\ell \in N_{j_\ell}$. Hence \begin{align*} x-y = \sum_{k=1}^{m} x_k + \sum_{\ell=1}^{n} (-y_\ell) \end{align*} is again a finite sum of elements drawn from the family $(N_i)_{i \in I}$. Therefore $x-y \in S$. For each $k$, since $N_{i_k}$ is a submodule of the left $R$-module $M$, it is closed under left scalar multiplication by elements of $R$, so $rx_k \in N_{i_k}$. By distributivity of scalar multiplication over finite sums in $M$, \begin{align*} rx = \sum_{k=1}^{m} rx_k. \end{align*} Thus $rx \in S$. Since $S$ is nonempty and is closed under differences and left scalar multiplication, $S \le M$.[/step]

[guided]The definition of the sum of a family of submodules is deliberately finite: an element of \begin{align*} S := \sum_{i \in I} N_i \end{align*} is not an arbitrary infinite series, but a finite sum of elements from the various $N_i$. The empty finite sum is allowed and is equal to $0$, so $0 \in S$. This proves the nonemptiness needed for the [submodule test](/theorems/8497). Now take arbitrary elements $x,y \in S$ and an arbitrary scalar $r \in R$. Because $x$ and $y$ lie in $S$, we may choose finite representations \begin{align*} x = \sum_{k=1}^{m} x_k \end{align*} and \begin{align*} y = \sum_{\ell=1}^{n} y_\ell, \end{align*} where $m,n$ are nonnegative integers, where $i_1,\dots,i_m,j_1,\dots,j_n \in I$, where $x_k \in N_{i_k}$ for each $k$, and where $y_\ell \in N_{j_\ell}$ for each $\ell$. If either finite list is empty, the corresponding sum is $0$. To prove closure under subtraction, we rewrite $x-y$ as one finite sum. Since each $N_{j_\ell}$ is a submodule, it is closed under additive inverses, so $-y_\ell \in N_{j_\ell}$ for every $\ell$. Therefore \begin{align*} x-y = \sum_{k=1}^{m} x_k + \sum_{\ell=1}^{n} (-y_\ell) \end{align*} is a finite sum whose summands still belong to members of the family $(N_i)_{i \in I}$. Hence $x-y \in S$. To prove closure under left scalar multiplication, use the module distributive law and the fact that each $N_{i_k}$ is closed under scalar multiplication. For every $k$, we have $x_k \in N_{i_k}$ and $N_{i_k} \le M$, so $rx_k \in N_{i_k}$. Thus \begin{align*} rx = \sum_{k=1}^{m} rx_k \end{align*} is again a finite sum of elements drawn from the same family. Hence $rx \in S$. Since $x,y \in S$ and $r \in R$ were arbitrary, $S$ is closed under differences and left scalar multiplication. Together with $0 \in S$, this proves $S \le M$.[/guided]

[step:Embed each summand as a one-term finite sum] Fix $i \in I$ and let $x \in N_i$. Then $x$ is the one-term finite sum \begin{align*} x = x \end{align*} with its only summand in $N_i$. Hence $x \in \sum_{j \in I} N_j$. Since $x \in N_i$ was arbitrary, $N_i \subset \sum_{j \in I} N_j$. [/step]

[step:Show every containing submodule contains the finite-sum set] Let $L \le M$ satisfy $N_i \subset L$ for every $i \in I$. Let $x \in \sum_{i \in I} N_i$. By definition, $x$ has a finite representation \begin{align*} x = \sum_{k=1}^{m} x_k \end{align*} where $m$ is a nonnegative integer, $i_1,\dots,i_m \in I$, and $x_k \in N_{i_k}$ for each $k$. Since $N_{i_k} \subset L$, each $x_k$ lies in $L$. Because $L$ is a submodule, it is closed under finite addition and contains $0$, so the finite sum $x$ lies in $L$. Therefore \begin{align*} \sum_{i \in I} N_i \subset L. \end{align*} This proves that $\sum_{i \in I}N_i$ is the least submodule of $M$ containing every $N_i$. [/step]