$\sum_{i\in I}\kappa_i$ vs. $\prod_{i\in I}\kappa_i$ Let $I$ be an arbitrary set and $(\kappa_i)_{i\in I}$ a family of cardinals.
What can be said about the relationship between
$$\sum_{i\in I}\kappa_i \qquad\text{and}\qquad  \prod_{i\in I}\kappa_i\qquad?$$
I know that for $I=2$, they are equal. But what if $I$ is an arbitrary set with at least two elements, does it always hold that
$$\sum_{i\in I}\kappa_i\leq\prod_{i\in I}\kappa_i$$
or even
$$\sum_{i\in I}\kappa_i=\prod_{i\in I}\kappa_i?$$
What else can be said?
 A: Let's write $S$ for the sum $\sum_{i\in I}\kappa_i$ and $P$ for the product $\prod_{i\in I} \kappa_i$.
If $I$ is empty, then $S = 0$ and $P = 1$, so $S<P$. From now on. let's assume $I$ is non-empty.
If some $\kappa_i$ is equal to $0$, then $P = 0$, so $P<S$ unless every $\kappa_i$ is equal to $0$, in which case $P = S$. From now on, let's assume all the $\kappa_i$ are non-zero.
Let $J = \{i\in I\mid \kappa_i > 1\}$ and $K = \{i\in I\mid \kappa_i = 1\}$, and let $\lambda = \sum_{i\in K}\kappa_i = |K|$. Then $S = \left(\sum_{i\in J}\kappa_i\right) + \lambda$ and $P = \prod_{i\in J}\kappa_i$. Note that we can have $P < S$ if $\lambda$ is large enough, i.e., if many of the $\kappa_i$ are equal to $1$. But in any case, we can reduce the question to comparing $\sum_{i\in J}\kappa_i$ and $\prod_{i\in J}\kappa_i$ and $\lambda$. From now on, let's assume all the $\kappa_i$ are greater than $1$.
If $I$ is finite and all the $\kappa_i$ are finite, the question is about finite arithmetic. If  $|I| = 1$, then $S = P$. If $|I|>1$, then since all the $\kappa_i$ are at least $2$, we have $S < P$, except in the single case $2+2 = 2\times 2$.
If $I$ is finite and at least one of the $\kappa_i$ is infinite, then $S = P = \text{max}_{i\in I} \kappa_i$. From now on, let's assume $I$ is infinite.
Based on all the above reductions, we're left with situation that $I$ is infinite and all the $\kappa_i$ are at least $2$. I claim that in this situation, $S\leq P$.
Let $\kappa = \sup_{i\in I} \kappa_i$. Then $S = \max(\kappa,|I|)$. First, we show that $|I|\leq P$, by finding an injective function from $I$ to $\prod_{i\in I} \kappa_i$. We map $i$ to the tuple in the cartesian product which is $1$ on the $i$-coordinate and $0$ on every other coordinate. This is well-defined because $\kappa_i \geq 2$ for all $i$.
Now if $\kappa\leq |I|$, then $S = |I|\leq P$. It remains to consider the case $|I| < \kappa$. Let $I_\infty = \{i\in I\mid \kappa_i \text{ is infinite}\}$.  Since $I$ is infinite, $\kappa > \aleph_0$, so $\kappa = \sup_{i\in I} \kappa_i = \sup_{i\in I_{\infty}} \kappa_i$.  Thus $S = \kappa = \sup_{i\in I_{\infty}} \kappa_i = \sum_{i\in I_{\infty}} \kappa_i$, since $|I_\infty|\leq |I|$. And since none of the $\kappa_i$ are  $0$, we have $\prod_{i\in I_{\infty}}\kappa_i \leq P$. So we have reduced to showing that $\sum_{i\in I_{\infty}} \kappa_i \leq \prod_{i\in I_{\infty}} \kappa_i$, i.e., that $S\leq P$ under the assumption that all of the $\kappa_i$ are infinite.
For each $i$, let $s_i\colon \kappa_i\to \kappa_i$ be the successor function $\alpha\mapsto \alpha+1$ ($s_i$ is well-defined since $\kappa_i$ is infinite). Then $s_i$ is injective and does not have $0$ in its range. We define an injective function  $\bigsqcup_{i\in I} \kappa_i \to  \prod_{i\in I} \kappa_i$ by mapping $\alpha\in \kappa_i$ to the tuple which is $s_i(\alpha)$ on the $i$-coordinate and $0$ on every other coordinate. This completes the proof.
But in this situation ($I$ is infinite and all the $\kappa_i$ are at least $2$), the question of whether $S < P$ or $S = P$ can be independent of ZFC. For example, we have:
\begin{align*}
\aleph_1 + \sum_{i\in \aleph_0} 2 &= \aleph_1 + \aleph_0 = \aleph_1\\
\aleph_1 \cdot \prod_{i\in \aleph_0} 2 &= \aleph_1 \cdot 2^{\aleph_0} = 2^{\aleph_0}.
\end{align*}
In this case, the question of whether the product is strictly larger than the sum is the continuum hypothesis!
I made lots of claims without proof above. If anything is not clear to you, let me know and I'll elaborate.
