# 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.

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Whenever I encountered these they represented 0 and 1 respectively, but there may be exceptions. What particular formula is this about ? – justt May 2 '13 at 22:30
@justt It isn't so much for a particular formula, but rather for when I write down nested sums/products. I'll edit a more concrete example into the question. – apnorton May 2 '13 at 22:33
@justt see example now... – apnorton May 2 '13 at 22:36
Does anybody have an example of $\sum_{k=a}^b$ or $\prod_{k=a}^b$ with $b<a-1$ from real life math literature? I think the case $b=a-1$ (meaning $\sum=0, \prod=1$) is quite common, but I wonder if the different possible interpretations for "even worse" cases are in practical use at all. – Hagen von Eitzen May 2 '13 at 22:44
@anorton The empty product in your example is common (and definitely equals $1$), not the product at the beginning of your question. – Did May 2 '13 at 22:46

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 $\prod\limits_{k=2}^{3}b_k=b_2b_3$.

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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.

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It depends. Some authors prefer to have $\sum_{k=a}^{b-1}+\sum_{k=b}^{c-1}$ to always eequal $\sum_{k=a}^{c-1}$ – Hagen von Eitzen May 2 '13 at 22:40