Interpretation of sums using $\cdots$ Consider the sum
$$\sum_{1\le k_1 < k_2 < \cdots < k_r \le n}k_1k_2\ldots k_r$$
Does this simply mean
$$\sum_{\substack{|K|=r}\\\inf(K)\ge 1\\\sup(K)\le n}\prod_{k\in K} k$$
I am specifically worried about the situation when $n=r=0$. In the second notation, this is clearly 1, but I'm not sure if it is 0 or 1 in the first notation. 
 A: I set out to write an answer that said that clearly the first notation gives $1$ as well but now I am not so sure anymore.
You are essentially defining the set $I = \{\,(k_1, \dots, k_r)\,\vert\,1 \leq k_1 < \dots < k_r \leq n\,\}$. I see three sensible ways to translate the condition $1 \leq k_1 < \dots < k_r \leq n$. 


*

*You can first split it into conditions for each $k_i$, i.e. $1 \leq k_1 < k_2$, $k_1 < k_2 < k_3$, $\cdots$, $k_{r-1} < k_r \leq n$ (and then split each of those into two inequalities). If you proceed like this, you don’t get any conditions in case $r = 0$, thus leaving you with the one-element set $\{()\}$ and a value of $1$.

*Alternatively, you translate the condition directly as $1 \leq k_1$, $k_1 < k_2$, $\cdots$, $k_r \leq n$ which would give you $1 \leq n$ in case that $r = 0$ and thus $I$ would be empty if $n = 0$ as well.

*You might also translate the condition as $1 \leq k_i$, $k_i \leq n$ for each $i$ and $k_1 < k_2$, $\cdots$, $k_{r-1} < k_r$. Treating the to relations on the outside differently doesn’t seem completely ridiculous, after all, it is a different relation. This would also give the value $1$.


It seems to me that the third interpretation is the correct one, but I really can’t give a better reason than that it agrees with your second notation which seems reasonable. If you want to use the first notation and the case $r = n = 0$ is important, you should probably mention which value you intend the sum to denote.
A: Usually the empty sum is defined to be $0$ and the empty product is defined to be $1$. 

We find in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik in section 2.1:
We write
  \begin{align*}
\sum_{P(k)}a_k\tag{1}
\end{align*}
  as an abbreviation for the sum of all terms $a_k$ such that $k$ is an integer satifying a given property $P(k)$. A property $P(k)$ is any statement about $k$ that can be either true of false. If $P(k)$ is false for all integers $k$, we have an empty sum; the value of an empty sum is defined to be zero.
Considering OPs sum
  \begin{align*}
\sum_{1\le k_1 < k_2 < \cdots < k_r \le n}k_1k_2\ldots k_r\tag{2}
\end{align*}
  we observe according to (1) that if $n=0$, the sum is zero, independent of the setting of $r$. We do not need to care about the specific structure of the summands, when the sum is already an empty sum.

The summands of $(2)$ are finite products
\begin{align*}
\prod_{j=1}^r k_j\tag{3}
\end{align*}
We find in section 4.2 that the value of an empty product is by definition given as $1$. So, if $r=0$ the product in (3) is equal to $1$.
A: The sum $\sum_{1\leq i < j \leq n}$ is merely a shorthand for $\sum_{1 \leq i \leq n}\sum_{i<j\leq n}$. Hence
$$\sum_{1 \leq k_1 <k_2 < \dots < k_r \leq n} k_1k_2\dots k_r = \sum_{1\leq k_1 \leq n}\sum_{k_1 < k_2 \leq n}\dots \sum_{k_{r-1} < k_r \leq n}k_1k_2\dots k_r$$
Written like this the interpretation when $r=0$ is obvious: There is no sum operator, so $n=0$ cannot produce an empty sum. It is then clear that the two expressions in the question are equivalent. 
