# A Summary Of Some Summery Summation

“Counting is

hard. This is what I keep saying.”

It all stemmed from an arithmetic series problem with a known sum, but an unknown number of terms. As these things are prone to do, it led to a quadratic equation; as those things are prone to do, that led to two possible values for $n$, one of which was negative.

This is fine; there is no problem here; you discard the negative root as impossible due to the context, collect three marks and move on to the next question. Unless, of course, you’re @realityminus3, in which case you ask “what does the negative root represent?”

That’s not a simple question. I found a simpler, related one: what is $\sum_{k=100}^{0} k$?

It’s a simpler question. However, it’s most definitely not a simple answer; I think there are reasonable arguments to be made for four distinct answers. Let’s start with the boring ones.

### Undefined

“It doesn’t make any sense to have the lower limit larger than the upper one” is, to me, a perfectly valid (if dull) way to read the notation. I don’t think anyone I asked made this argument, but it’d be a bit off to ignore it.

### 0

If you’re going to say something about the empty set, I will fight you.

I’m afraid Christian is going to have to take this one up with [Wolfram|Alpha](http://www.wolframalpha.com/input/?i=sum(k%3D100..0,+k) and Wikipedia, if they think they’re hard enough.

Wikipedia, in particular, explicitly defines this as 0 - the logic being that this notation means $\sum_{100\le k \le 0} k$, which is indeed the empty set.

### 5050

This is by far the preferred answer among the Twitterers who responded to a poll on the subject (as if maths is decided by popular opinion!):

Is $\sum_{r=100}^{1} r$:

[Aside: You’ll note that I erroneously put -5050 in the poll instead of -4950. This was not an attempt to invalidate the poll and force a rerun, honestly.]

Fully 82% of those 41 people chose the ‘obvious’ answer that summation doesn’t care about order.

Some of the arguments include:

**@adamcreen**: “Substitute and add. Discrete so not area so can’t be negative.”**@nextlevelmaths**: “This is silly, really, because with addition being commutative you’d just never have reason to write that, other than to confuse.”**@mathteacher1729**: “Worth noting that when I see it in AP Calculus, the smaller number is typically the subscript: $ \sum_{r=1}^{100} r $.”**@graememcrae**: “The integral from 100.5 to 0.5 of r dr is -5050, but the sum from 100 to 1 step -1 of r is 5050. Reason: integral $\approx$ sum × step.”**@topometallo**(on Mathstodon): “Summation doesn’t care about order. If you think -4950 because of the ‘step -1 reason’ why should the first ‘100’ become negative?”

Now, I have a dilemma: do I, @peterrowlett-like, stay out of the argument altogether, or get into commentary on these suggestions? I think, for once, I’ll leave it. For now, at least.

### -4950

The analysts on Mathstodon - notably @alephthought, @jeffgerickson and @byorgey ((Brent considers himself a combinatorialist rather than an analyst, but he probably also says po-tah-to.)) - note that summation has, for $a<b<c$, a really useful property: $\sum_{r=a}^b f(r) + \sum_{r=b+1}^c f(r) = \sum_{r=a}^c f(r)$. That would be a nice property to have *without* the restriction, in which case $\sum_{r=0}^{99} r + \sum_{r=100}^{0} r = \sum_{r=0}^{0} r = 0$. The first term on the left is 4950, so the second must be -4950.

This was spiritually the same as my (piratical) argument: when you’re integrating, flipping the limits negates the integral, so something similar ought to work for summation.

But if you think that’s piratical, wait until you hear what Barney Maunder-Taylor has to say. Inspired by a Year 6 question, he pondered ‘what happens if the numbers join up around the back?’ This idea suggests that $\sum_{r=100}^{0} r = \sum_{r=100}^\infty r + \sum_{r=-\infty}^{0} r$. Of course, neither of those sums converge, but many of their individual terms correspond each other. In particular, you’d expect $\sum_{r=100}^\infty r + \sum_{r=-\infty}^{-100} r$ to be 0. All you’re left with is $\sum_{r=-99}^{0} r$, which does indeed give -4950.

A final argument, based on this particular summation, is that $\sum_{r=a}^{b} r = \int_{r=a-\frac {1}{2}}^{b+\frac{1}{2}} r \d r$, if $a \le b$. If we remove the restriction, $\sum_{r=100}^{0} r = \int_{99.5}^{0.5} r \d r = \left[ \frac{1}{2}r^2\right]_{99.5}^{0.5} = -4950$.

### Summing up

Personally, I’m much more convinced by the arguments for -4950 (for convenience) and 0 (by definition) than for 5050 (by simplicity) or for undefined (by dodging the question). I’ll leave the last word to @xander (on Mathstodon), who says: “Regard the sum as a Lebesgue integral with respect to counting measure on the integers, and adopt the convention that orientation matters (that is, $\int_b^a f =−\int_a^b f$). Of course, any of the other three answers could be justified, as well. It depends on what you mean by $\sum$.”

I’d be interested to hear different arguments and/or explanations in the comments below!

* Thanks to everyone who took part in the discussion! * Edited 2017-06-05 to correct a couple of typos. I wondered who’d be the first to spot them, and it turned out to be @realityminus3. Thanks! * Edited 2017-06-05 to give credit to Jeff Erickson, whose toot had got lost among the favourites. * Edited 2017-06-05 to clarify Brent Yorgey’s field of expertise.