Summation and Product Bounds If I have a sum or product whose upper index is less than its start index, how is this interpreted?  For example:
$$\sum_{k=2}^0a_k,\qquad \prod_{k=3}^1b_k$$
I want to say that they are equal to the empty sum and empty product, respectively, but I don't know.
(This question arises from seeking shortened forms for denoting some nested series/sequences, where the upper index of the inner sum/product is the variable for the outer sum/product.)
Example for why I was wondering:
$$\sum_{n=0}^\infty\left[\frac{\prod_{k=0}^{n-1}\left(4k-1\right)}{(2n+1)!}\right]$$
Note that, for the first case of $n=0$, the product is in a situation like I describe.
 A: A common interpretation of these somewhat unorthodox summation signs is based on the observation that $\sum\limits_{k=i}^{n+1}a_k=a_{n+1}+\sum\limits_{k=i}^{n}a_k$ for every $n\geqslant i$. Extending this, one should expect (and indeed this is the most widely used convention) that $\sum\limits_{k=i}^{i-1}a_k=0$ for every $i$ and every sequence $(a_n)$. Likewise, $\prod\limits_{k=i}^{i-1}b_k=1$ for every $i$ and every sequence $(b_n)$.
The summation $\sum\limits_{k=2}^{0}$ goes one step further but logic would suggest that $\sum\limits_{k=2}^{0}a_k=-a_1$, although I must confess having never met such uses.
Likewise, one should probably consider that $\prod\limits_{k=3}^{1}b_k=1/b_2$ for every nonzero $b_1$, $b_2$, $b_3$, as can be deduced from the identity $\prod\limits_{k=3}^{1}b_k\cdot\prod\limits_{k=2}^{3}b_k=\prod\limits_{k=3}^{3}b_k=b_3$ and from the fact that $\prod\limits_{k=2}^{3}b_k=b_2b_3$.
A: What you are really doing with $\sum$ when you see
$$
\sum_{k\in\mathcal{A}}a_{k}
$$
is adding every element in $\mathcal{A}$ to the additive identity $0$. Likewise, when you see
$$
\prod_{k\in\mathcal{A}}a_{k},
$$
you are multiplying every element in $\mathcal{A}$ to the multiplicative identity $1$.
Note: the notation with a subscript $k=a$ and superscript $k=b$ is just sugar for $\mathcal{A}=\left\{a,a+1,\ldots,b\right\}$ whenever $a\geq b$ and $\mathcal{A}=\emptyset$ otherwise.
