# A little curiosity about sum notation

In summation notation($$\sum$$), can the stopping point be smaller than the starting point?
For example, can I say $$\sum_{i=1}^0 i = 0$$
because $$\ 1 > 0$$ so it does not sum anything??

• The stopping and starting point tell us the number of terms to be added. Commented Apr 27, 2020 at 10:42
• I guess this would be invalid and the correct way of writing it is $\sum_{i=0}^1 i = 1$. Commented Apr 27, 2020 at 10:42
• This is fine. It is an "empty sum", so it is equal to zero. Commented Apr 27, 2020 at 10:43
• I've not seen it, but I can think of at least three ways to interpret that (the sum being $-1$, $0$ or $1$) so I think you shouldn't write that unless you explain what you mean. Commented Apr 27, 2020 at 10:44

Bear in mind $$\sum_{i=a}^bf(i)$$ is shorthand for $$\sum_{i\in S}f(i)$$ with $$S:=\{i\in\Bbb Z|a\le i\le b\}$$, so your notation is a special case of an empty sum, as would be $$\sum_{i=1}^{-1}i$$. In general, $$\sum_{i=a}^bf(i)$$ sums $$\max(b-a+1,\,0)$$ elements (you can also denote this $$(b-a+1)^+$$). You can get empty products with the same rules, just replacing $$\sum$$ with $$\prod$$.

• Thank you! Then will the expression like $\sum_{i=1}^{9-k} i!S(9-k, i)$ be equal to $\ 0$ if $\ k = 9$? Commented Apr 27, 2020 at 10:55
• @RyanRo An awfully specific example, but yes.
– J.G.
Commented Apr 27, 2020 at 10:56

Wikipedia gives the following formal definition:

\begin{align} \sum_{i=a}^bg(i)&=0,\text{ for }b\lt a\\ \sum_{i=a}^bg(i)&=g(b)+\sum_{i=a}^{b-1}g(i),\text{ for }b\ge a. \end{align}

Remark: It's somewhat unclear (to me at least) whether Wikipedia is tacitly assuming that $$a,b\in\mathbb{Z}$$ in its recursive definition. This could be important because you sometimes see people write things like

$$\sum_{i=0}^\sqrt ng(i)$$

where they (probably) mean

$$\sum_{i=0}^{\lfloor\sqrt n\rfloor}g(i)$$