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I have a sum defined as $ \sum_{\mu=0}^{k-1} f(\mu) $ , what does it mean in the case that $k=0$? Then I have a sum running from $0$ to $-1$ which I suppose doesn't make sense? Wolfram alpha gives $0$ no matter what I put in for $f(\mu)$, why is this the case?

I think I have a better understanding of what this means in the context of integrals, but summations not so much. Is there a connection?

calvingentry
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  • If you have $\sum_{i=a}^b$ and $b<a$, convention is just that the sum is zero. It's a bit different with integration since the measure is also involved. – Chappers Jul 14 '17 at 17:30

1 Answers1

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It's a matter of convention. The "empty" or "vacuous" sum is defined to be zero:

$$\sum_{\text{false}} x := 0$$

no matter what $x$ is.

Similarly, the "vacuous product" is defined to be one:

$$\prod_{\text{false}} x := 1$$

no matter what $x$ is.

Here the "$\text{false}$" is $0 \le x \le -1$.

This is a definition rather than a theorem, but it is consistent with results and that use the summation notation.

GFauxPas
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