Is \( \sum_{r=100}^{0} r \):

a) 5050
b) 0
c) -5050
d) other

I have an Opinion, but I don't want to share it until I know whether it's controversial (or worse yet, Wrong.)

@icecolbeveridge if you're going to say something about the empty set, I will fight you

@christianp I have, appropriately, nothing to say about the empty set ;o) Thanks for response, this may become a blog post.

@icecolbeveridge I'm going to go with (c). If you want (sum from a to b) + (sum from b to c) = (sum from a to c) to hold in all cases, you want c to be the answer, just like for integrals. A sum is just an integral with respect to point mass measure anyway. (You can tell I'm an analyst.)

@icecolbeveridge [a] Summation doesn't care of order. If you think -5050 cause the "step -1 reason" why should the first "100" become negative?

@icecolbeveridge It depends. With my computer scientist hat on, I say (b); it's a silly way to express the sum of the empty set. With my combinatorialist hat on, I say (c); if we want to usefully think of \( \sum \) as a discrete analogue of integration, then the sign ought to reverse when you reverse the limits of summation.

@icecolbeveridge I like (c). Regard the sum as a Lebesgue integral with respect to counting measure on the integers, and adopt the convention that orientation matters (that is, \(\int_a^b f = -\int_b^a f\)). Of course, any of the other three answers could be justified, as well. It depends on what you mean by \(\sum\).

@icecolbeveridge Either (b) 0 or (d) -4950.

Case for (b): You're summing over all integers \(r\) such that \(100\le r\le 0\).

Case for (d): We want the identity \(\sum_{i=a}^b + \sum_{i=b+1}^c = \sum_{i=a}^c\) to hold for all \(a,b,c\). But notice the +1 in the second sum! So \( \sum_{r=0}^{99} r + \sum_{r=100}^{0} r = \sum_{r=0}^0 r = 0 \)

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