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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??

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  • $\begingroup$ The stopping and starting point tell us the number of terms to be added. $\endgroup$
    – Learning
    Commented Apr 27, 2020 at 10:42
  • $\begingroup$ I guess this would be invalid and the correct way of writing it is $\sum_{i=0}^1 i = 1$. $\endgroup$
    – Vishu
    Commented Apr 27, 2020 at 10:42
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    $\begingroup$ This is fine. It is an "empty sum", so it is equal to zero. $\endgroup$
    – wormram
    Commented Apr 27, 2020 at 10:43
  • $\begingroup$ 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. $\endgroup$
    – skyking
    Commented Apr 27, 2020 at 10:44

2 Answers 2

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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 $\prod$.

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

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

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