I tend to believe that mathematical results which seem surprising or paradoxical are better interpreted as just being bad results—and they are really indicators of places that our foundations have gone wrong. For instance I think Banach-Tarski is an indictment of measure theory, not a surprising fact about the universe; the long list of pathological counterexamples in topology is an argument for switching to a finistic foundation which routes around all of them; and the surprising implications of the axiom of choice are evidence that we should discard set theory as a mathematical foundation. That sort of thing. (Unsurprisingly, the list of disagreeable results is almost entirely statements involving infinity.) I don’t think I can convincingly argue for these claims yet, but I do believe them.

A particularly salient example of the problem is found in the summation of divergent series. I have long held a vendetta against all those online math explainers who gave credence to nonsensical manipulations like the infamous \(1 + 2 + 3 + 4 + \ldots \? -\frac{1}{12}\) sum, which I wrote about in one of my first articles. That post was inspired by a Numberphile video which had a very disappointing discourse around it at the time: people were far too willing claim that the sum equaling \(-1/12\) was true in some sense, or even literally true, instead of what seems like to me like the obviously correct conclusion: that math has an unresolved hole in it, which any layperson can clearly see but which people trapped inside the ‘framework’ of modern mathematics somehow miss. Considerations of questions like this have led me personally to a version of finitism (although not a version I have seen anyone else exactly ascribed to). But that is for another time.

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Anyway this post is something of a followup to that earlier one. In the intervening decade I have come to know a much more simple and pleasing exposition for those divergent sums than I knew at the time, one which I think dispels all possible objections. I thought I would write it out as a standalone article in case anyone else wants to see this laid to rest (or, I guess, thinks there’s something I’m still missing). Nothing in here is particularly deep, and that’s kind of the point. It is just a simple explanation for a simple thing. Unfortunately it is somewhat dry to explain all in detail… I imagine that it will be interesting if you specifically care about this particular puzzle, and not if you don’t, and that’s fine.

(Parts of this article were previously part of an earlier post about Taylor series. This section was much more likely to be interesting to anyone else than the rest so I decided to split it out, and have added some more material in the process.)

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1. The Example of \(1/(1-x)\)

First we will discuss the objectively-coolest Taylor series, that of

\[f(x) = \frac{1}{1-x}\]

which is

\[S(x) = 1 + x + x^2 + x^3 + \ldots\]

This converges to the true value of \(f(x)\) for \(\| x \| < 1\). For example,

\[\begin{aligned} f(\frac{1}{2}) &= \frac{1}{1-\frac{1}{2}} = 2 \\ S(\frac{1}{2}) &= 1 + \frac{1}{2} + \frac{1}{4} + \ldots = 2 \end{aligned}\]

The equality holds specifically if the \(1+1/2+1/4+\ldots\) is regarded as a limit, which is standard: it’s equivalent to how \(1.99\overline{9} = 2\). In fact it’s the same phenomenon exactly, since the statement \(1+\frac{1}{2}+\frac{1}{4} + \ldots = 2\) becomes \(1.11\overline{1} = 10\) when written in binary. However, keep in mind for later that “regarded as a limit” is not the only way to interpret equality; it’s just the one we’re steeped in and therefore used to.

We can also write down a Taylor series for \(f(x)\) around points other than \(x=0\). A useful one is the series around \(x=\infty\), which is the same thing as a series in the variable \(1/x\) around \(1/x=0\). I’ll write this as \(S_{\infty}(x)\) to distinguish it from the other one:

\[\begin{aligned} f(x) &= \frac{1}{1-x} \\ &= (-\frac{1}{x}) \frac{1}{1 - \frac{1}{x}} \\ &= -\frac{1}{x} [1 + \frac{1}{x} + \frac{1}{x^2} + \ldots] \\[0.5em] S_{\infty}(x)&= -\frac{1}{x} - \frac{1}{x^2} - \frac{1}{x^3} - \ldots \end{aligned}\]

This converges for \(\| x \| > 1\), for example

\[\begin{aligned} f(2) &= \frac{1}{1-2} = -\frac{1}{2} \frac{1}{1 - \frac{1}{2}} = -1 \\ S_{\infty}(2) &= -\frac{1}{2}[1 + \frac{1}{2} + \frac{1}{4} + \ldots] = -1 \end{aligned}\]

There are also series around any \(x = k \neq 1\), but we won’t need tohs.e

Outside the radii of convergence each of these series gives an infinite result which is not equal to \(f(x)\). For instance \(S(x)\) converges for \(\| S \| \leq 1\), and outside of that it diverges to infinity:

\[S(2) = 1 + 2 + 4 + 8 + \ldots\]

The paradox arises because, although this value is infinite, there appears to be a sense in which it does seem to encode the value of \(f(2) = 1/(1-2) = -1\) somehow: when this divergent sum is evaluated by any of the various divergent series summation techniques, it gives the right answer! For the example the naivest way to evaluate a divergent sum is to algebraically manipulate it to be finite again:

\[\begin{aligned} S(2) &= 1 + 2 + 4 + 8 + \ldots \\ S(2) - 2 S(2) &= [1 + 2 + 4 + \ldots] - (2 + 4 + 8 + \ldots) \\ (1-2)S(2) &= 1 \\ S(2) &= -1 \end{aligned}\]

We are left with the puzzle of explaining how this is even possible.

One might interpret this to mean that, although the result of \(S(2)\) is not a number, it still contains the data \(-1\) somehow—maybe in a form like \(-1 + O(\infty)\)? I have seen this type of explanation a lot. People freqeuntly link Tao’s article on the subject, which shows how you can “smooth out” these sums to remove the jumps at each discrete integer. Doing this typically reveals a constant term which has the finite value that the summation techniques get. The implication is that the sum \(S(2) = 1+2+4+8+\ldots\) contains more data in than just its ‘value’ \(\infty\); in particular it encodes its growth rate, which has a finite part equal to \(-1\).

I’m sure that is true, but I’ve come to think that it’s making things too complicated. Also, it doesn’t really explain what is going on. Do divergent sums have values, or not? If they’re not numbers, then what are they, and what algebra do they obey? If this value \((-1)\) is in there, what else is in there?

So now I will describe a very simple way of thinking about this that is completely satisfactory for intuition. Depending on your background it might seem a bit elementary, so I feel almost weird making a big deal about it, but I think it’s important for staying sane. We are simply going to refuse to consider “infinite” sums at all and only allow ourselves to regard them finitistically, up to some unspecified position \(N\), and it turns out this explains everything.

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2. Exact Divergent Sums

Recall that for any finite \(N\), the partial sum

\[S_N(x)= 1 + x + x^2 + \ldots + x^N\]

is not a multiplicative inverse of \(1/(1-x)\). If it were we would have \((1-x) S_N(x) = 1\), but we don’t. Instead we have

\[(1-x) S_N(x) = (1-x)(1 + x + x^2 + \ldots + x^N) = 1 - x^{N+1}\]

Evidently the real value of \(S_N(x)\) is

\[S_N(x) = \frac{1-x^{N+1}}{1-x} = \frac{1}{1-x} - \frac{x^{N+1}}{1-x}\]

Normally we think of the value of \(N\) as implicitly going to infinity, and then it’s clear why this converges only for \(\| x \|<1\), because that’s the case where the error \(x^{N}\) goes to \(0\). This is why \(S(1/2) = 2\) is correct, because \((1/2)^N\) becomes indistinguishable from \(0\) for large enough \(N\). And the reason that \(S_N(2) = -1\) diverges is that

\[S_N(2) = \frac{1}{1-2} - \frac{2^{N+1}}{1-2} = 2^{N+1} - 1\]

is not the value of \(\frac{1}{1-2}\), even though it is indeed the value of \((1+2+4+\ldots + 2^N)\).

When we find that \(S_N(2) \? -1\) under divergent summation techniques it’s because the algebraic manipulations used are somehow erasing the \(x^{N+1}\) term even though it should contribute to the result. Specifically it happens at the step where we subtracted the series from itself. It should have read:

\[\begin{aligned} (1-x)S_N(x) &= (1 + x + \ldots + x^N) \\ &\; \;\;\; - (x + x^2 + \ldots + x^{N+1}) \\ 1 &\stackrel{??}{\neq} 1 - x^{N+1} \\ \end{aligned}\]

Writing this as \(1\) requires arbitrarily dropping the \(O(x^N)\) term even though it is not necessarily small—but since it was hiding behind the ellipses we “got away with it”. In fact the computation that gave \(S(2)=-1\) should have read

\[\begin{aligned} (1-2) S_N(2) &= S_N (2) - 2S_N(2) \\ -S_N(2) &= -1 + 2^{N+1} - 2(-1 + 2^{N+1}) \\ &= 1 - 2^{N+1} \\ S_N(2) &= 2^{N+1} - 1 \end{aligned}\]

which is correct for all \(N\). When we wrote that \(S(2) = -1\), it’s not that the \(2^{N+1}\) term actually cancelled—it’s just that we dropped it. Which was aided by the fact that writing it as \(1+x+x^2+x^3+\ldots\) notation made it easy to do so. So regardless of what confusions you might have about \(S(x)\), there is no mystery with \(S_N(x)\), and one way of looking at the problem is just that we found some algebraic manipulations which allow one to pretend that the \(2^{N+1}\) terms cancel out when they really don’t.

My takeaway is that we ought to insist that our infinite sums are really partial sums: \(S(x)\) doesn’t need to have a value at \(N \ra \infty\), just at some finite \(N\) that suits our purposes. It’s fine if we want to omit it from our notation but we don’t get to also complain when that leads to confusing algebra, because the \(N\)-dependent term was really supposed to be there the whole time.

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Just in case it still seems like you might generally be able to figure out what the constant terms of a series are without seeing their \(N\)-dependencies, here is a more complicated example which can only be resolved by including the \(x^{N+1}\) terms.

Consider the function \(g(x) = \frac{1+x}{1+x+x^2}\) which has \(g(1) = 2/3\). The summation for \(g(1)\) appears to give an unsummable divergent series that looks exactly like \(S(-1)\):

\[\begin{aligned} g(x) &= \frac{1+x}{1+x+x^2} \\ &= \frac{1-x^2}{1-x^3} \\ &= (1-x^2)(1 + x^3 + x^6 + \ldots) \\ &= 1 - x^2 + x^3 - x^5 + x^6 - x^8 + \ldots \\ g(1) &= 1 - 1 + 1 - 1 - 1 + 1 + \ldots \end{aligned}\]

The manipulation \(\frac{1+x}{1+x+x^2} = \frac{1-x^2}{1-x^3}\) is true if you regard the value at \(x=1\) as a limit, whereuon it has the value \(g(1) = 2/3\) by l’Hôpital’s rule. But there’s no way to get \(\frac{2}{3}\) from the series \(1-1+1-1\ldots\) alone—clearly, because it is the same series as \(S(-1)\), which we have to suppose equals \(f(-1) = 1/2\) if we are buying the divergent sum techniques. So how are we to argue that \(g(1) = 2/3\)? The ‘data’ for that, if it is found in the series at all, must be encoded in the powers of \(x\)—the \((+x^0 -x^2 +x^3 -x^5 +x^6 \ldots)\) pattern. But there’s no way to say that it is found in \(1-1+1-1\ldots\) as well. What to do?

The mystery gets resolved if we also look at the \(O(x^N)\) term:

\[\begin{aligned} \frac{1-x^2}{1-x^3} &= (1-x^2)(1 + x^3 + x^6 + \cdots + x^{3N} + \frac{(x^3)^{N+1}}{1-x^3}) \\ &= 1 - x^2 + x^3 - x^5 + \cdots + x^{3N} - x^{3N+2} + x^{3N + 3} \frac{1-x^2}{1-x^3} \\ &\stackrel{x=1}{\Ra}(1-1) + (1-1) + \cdots + (1-1) + \frac{2}{3} \end{aligned}\]

This time the remainder term which is otherwise dropped is doing all the work: every other pair of terms cancel, and then the remainder gives the full value \(2/3\). So in this case erasing the \(N\)-dependence would leave something unusable. (You can also get a version of \(1-1+1-1+\ldots\) which equals any rational \(m/n\) you want by computing \((1-x^m)/(1-x^n)\) instead.)

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So that is how I think one can stay sane thinking about divergent sums. They were never divergent at all. Their algebraic properties follow entirely from treating them as partial sums to an arbitrary \(N\) and just making sure to be careful about tracking it.

This is an trick that turns out to work in a lot of settings, by the way: any time an infinite limit does something odd, just stop regarding it as infinite and do everything with a finite range instead, but keep the range as a free variable in the expression. Usually you will figure out what is going on pretty quickly.

One might object that this approach is really just a way of skirting around the issue. Wasn’t the actual mystery about the limit as \(N \ra \infty\)? After all that is the thing that has \(S(1/2) = 2\) and seems to also allo for \(S(1) = -1\). The mystery of divergent sums is in how they seem to work after taking limits, despite the limits being invalid—I hear you say.

I don’t buy it. I really think the limits don’t actually matter. What makes \(1+1/2 +1/4+\ldots = 2\) is that it is indistinguishable from \(2\) in a practical sense, not that they are literally equal. This is what is meant by the epsilon-delta definition of a limit anyway: for any accuracy (epsilon) you want, I can show a value that is close enough to \(1/2\) (by delta) to get that close to \(2\) (by epsilon). It was never needed that they be literally equal. It’s that they are equal enough that you can’t tell.

This is the ‘finitist’ perspective which I like to adopt anytime math does something weird, because it cuts through through all the mystery: you can’t complain about limits at infinity doing mysterious things if you refuse to take any limits at all. Although math traditionally assumes limits and infinitism everywhere, physical applications necessarily do not care about the difference. In a physical theory, any epsilon-delta limit you’re using doesn’t actually have to hold for all choices of \(\e\), but rather only for the choices of \(\e\) you actually have—which are finite values, characterizing the ‘resolution’ with which you can measure the number. Therefore one does not need to really consider limits, except insofar as they summarize the approximation process succinctly. If they start giving funny answers, the right thing to do is to make your \(N\)-dependencies explicit again—hopefully you did not forget where they are—and keep going. Which is not the usual way that limits are conceptualized, at least in pure mathematics, but I think it’s how they should be viewed.

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3. Some other sums

\(S(x) = 1+x+x^2 +\ldots\) is, admittedly, the easiest divergent series to understand. Therefore, to demonstrate that this way of thinking works more generally, and also mostly just to have a record of these calculations, I will go through each of the other interesting divergent sums I know and show how this ‘finitistic’ perspective gives the right answer. (Well, most of them. Unfortunately the sums involving primes are beyond me because I have never managed to absorb the content of number theory.)

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The first series to consider is (wiki):

\[P = 1 + 1 + 1 + 1 + \ldots\]

Which happens to be \(S(1)\). An easy manipulation shows that \(P\) equals… any integer you want?—

\[\begin{aligned} P - P &= 1 + 1 + 1 + 1 + \ldots \\ &\;\;\;\;\;\,-(1 + 1 + 1 + 1 + \ldots)\\ &= 1 \\ &= 1 + 1 + 1 + 1 + \ldots \\ &\;\;\;\;\;\;\;\;\;\;\;\;-( 1 + 1 + 1 + 1 + \ldots) \\ &= 2\ \end{aligned}\]

—which can’t be right.

If we fix \(P\) to have a finite number of terms \(N\) then its sum is clearly \(P = N\) To handle this better, we will label the sum with a subscript to indicate by how many terms were included in it, like this:¹

\[P_N = S_{N-1}(1) = \underbrace{1+1+1+1+\ldots}_{N \text{ times}} = \sum_1^N 1 = N\]

Then we can say concretely:

\[\begin{aligned} P_N - P_N &= 0 \\ P_N - P_M &= N - M \end{aligned}\]

Which dispels the mystery. You can get any number if you use two un-“correlated” copies of the value, but if you subtract the same value from itself you of course get \(0\), regardless of how you try to arrange the terms. So this just works and is basically trivial.

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Next we have Grandi’s Series, which is \(S(-1)\):

\[Q = 1 - 1 + 1 - 1 + \ldots\]

It is ‘summable’ with the classic manipulation

\[\begin{aligned} Q + Q &= 1 - 1 + 1 - 1 + \ldots \\ &\;\;\;\;\;\,+(1 - 1 + 1 - 1 + \ldots) \\ &= 1 \\ \Ra Q &\? \frac{1}{2} \\ \end{aligned}\]

The actual value is clearly

\[\begin{aligned} Q_N &= S_{N-1}(-1) = \frac{1}{2}(1 - (-1)^N) \end{aligned}\]

If we repeat the calculation with the actual value,

\[Q_N + Q_N = 2Q_N = 1 - (-1)^N\]

it is clear that the \(N\)-dependency was inadvertently being dropped.

However, we saw before that \((1-x^2)/(1-x^3)\) also gives the same series, yet equals \(\frac{2}{3}\). How do we reconcile this? For that matter, how does a sum of \(\pm 1\)s ever equal a non-integer? Why, it is simple. Remember the actual equivalency was

\[\begin{aligned} \frac{1-x^2}{1-x^3} &= (1-x^2)(1 + x^3 + x^6 + \cdots + x^{3N} + \frac{(x^3)^{N+1}}{1-x^3}) \\ &= 1 - x^2 + x^3 - x^5 + \cdots + x^{3N} - x^{3N+2} + x^{3N + 3} \frac{1-x^2}{1-x^3} \\ &\stackrel{x=1}{\Ra} Q_{2N} + \frac{2}{3} \end{aligned}\]

So it is not the case that \((1-x^2)/(1-x^3)\) equals the sum \(1-1+1-1+\ldots\). Instead it has an additional term at the end, whose value is exactly \(2/3\). Whew. No “sum of integers equals a fraction” today.

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Next up,

\[R = 1 - 2 + 3 - 4 + \ldots\]

(wiki) with the classic manipulation

\[\begin{aligned} R + R &= 1 - 2 + 3 - 4 + \ldots \\ &\;\;\;\;\;\,+(1 - 2 + 3 - 4 + \ldots) \\ &= 1 -1 + 1 - 1 + \ldots \\ &= Q \? \frac{1}{2} \\ \Ra R &\? \frac{1}{4} \end{aligned}\]

We already know the value of \(Q\) is wrong there, and probably so is the value of \(R\), since the \(N\) dependenceis are missing. How do we do it right? Well, the partial sums of \(R\) are

\[(1, -1, 2, -2, 3, -3, \ldots)\]

It is a bit harder to see how to write this as a closed form. It is almost \((1, \frac{3}{2}, 2, \frac{5}{2}, \ldots)\), except that every other term is wrong by a certain amount. Once you rewrite it like this

\[(\frac{1}{4} + \frac{3}{4}, \frac{1}{4} - \frac{5}{4}, \frac{1}{4} + \frac{7}{4}, \frac{1}{4} - \frac{9}{4}, \frac{1}{4} + \frac{11}{4}, \frac{1}{4} - \frac{13}{4}, \ldots)\]

the closed form becomes obvious:

\[R_N = \frac{1}{4} - (-1)^N \frac{2N+1}{4}\]

This is why the invalid computation that canceled out the \(N\)-dependency gave \(R = 1/4\): it was just dropping the \(N\)-dependence (again).

Another hacky way of getting to \(R=1/4\) that I’ve seen around is to square \(Q\) and then rearrange the terms in a certain way:

\[\begin{aligned} Q^2 &= (1-1+1-1+\ldots)(1-1+1-1+\ldots) \\ (\frac{1}{2})^2 &= 1 - 1 + 1 - 1 + \ldots \\ & \;\;\;\;\;\,-(1-1+1-1+\ldots) \\ & \;\;\;\;\;\;\;\;\;\;\;\;+(1-1+1-1+\ldots) \\ & \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;- \text{(...etc...)} \\ \frac{1}{4} &\? 1 - 2 + 3 - 4 + \ldots\\ \end{aligned}\]

For getting \(R_N\) this obviously won’t do at all: once we switch to actually tracking the length of the partial sums there are going to be a bunch of floating \(\pm 1\)s at the end of each term which don’t accumulate to give a term in \(R_N\). However, we can fix this method. If it is going to compute a particular value of \(R_N\), the sum should actually be

\[R_N = Q_N - Q_{N-1} + Q_{N-2}-Q_{N-3} \ldots\]

such that every sum terminates on exactly the \(N\)th term of \(1-2+3-4 +\ldots\). Since \(Q_N\) is \(0\) for \(N\) even and \(1\) for \(N\) odd, it simplifies to

\[\begin{aligned} R_N = \underbrace{Q_N - Q_{N-1} + Q_{N-2}-Q_{N-3} + \ldots \mp Q_1}_{N \text{ terms}} &= \begin{Bmatrix} -\frac{N}{2} & N \text{ even} \\[0.5em] \frac{N+1}{2} & N \text{ odd} \end{Bmatrix} \end{aligned}\]

Which happens to be another way of writing \(R_N = \frac{1}{4} - (-1)^N \frac{2N+1}{4}\). So that’s good. The reason \(Q^2 \? R \? \frac{1}{4}\) seemed to work is because it is equivalent to the constant terms of this calculation: naturally all the \(Q_{N-K}\)s have constant term \(1/2\), and so did the number of terms in the sum since it oscillated between even and odd numbers.

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4. -1/12

Finally we have the Ramunajan sum

\[T = 1 + 2 + 3 + 4 + \ldots\]

Which can be ‘summed’ by reducing to the \(R = \frac{1}{3}\) from above:

\[\begin{aligned} T - 4T &= 1 + 2 + 3 + 4 + 5 + 6 + \ldots \\ &\;\;\;\,- 4(1 \;\;\;+ \;\;\; 2 \;\;\;+\;\;\; 3 +\ldots) \\ &= 1 - 2 + 3 -4 + 5 -6 \\ -3 T &= R \\ &\? \frac{1}{4} \\ T &= - \frac{1}{12} \end{aligned}\]

This one uses a new trick, of ‘spacing out’ the series in order to act on every other term in the sum, which we will have to handle. There is also another problem, which is that we know the closed form for the partial sums for this series—it’s

\[\sum_{k=1}^N k = \frac{N(N+1)}{2}\]

which decidedly does not contain a \(-\frac{1}{12}\) term. So where does that value come from?

To get a closed form for \(T_N\) we will attempt to duplicate the summation logic while carefully tracking the length of the partial sums. The ‘spacing out’ step requires including only the terms up to exactly the \(N\)th term of \(T\). For this we need to use \(T_{N/2}\) if \(N\) is even or \(T_{(N-1)/2}\) if \(N\) is odd, like this:

\[\begin{aligned} R_N &= \begin{cases} T_N - 4 T_{N/2} & N \text{ even} \\ T_N - 4 T_{(N-1)/2} & N \text{ odd} \\ \end{cases} \end{aligned}\]

More concisely:

\[R_n = T_N - 4 T_{\lfloor N/2 \rfloor}\]

Using \(R_N = \frac{1}{4} - (-1)^N \frac{2N+1}{4}\) we can rework this into a recurrence for \(T_N\):

\[T_N = \frac{1}{4} - (-1)^N \frac{2N+1}{4} + 4 T_{\lfloor N/2 \rfloor}\]

(with \(T_0 = 0\)). It is not too hard to check that \(T_N = \frac{N(N+1)}{2}\) obeys this relation: if \(N\) is even then

\[\begin{aligned} \frac{N(N+1)}{2} &= \frac{N^2}{2} + \frac{N}{2} \\ &= -\frac{N}{2} + (N + \frac{N^2}{2}) \\ &= \underbrace{(\frac{1}{4} -\frac{2N+1}{4})}_{R_N} + 4(\underbrace{\frac{(N/2)(N/2+1)}{2}}_{T_{\lfloor N/2 \rfloor}}) \end{aligned}\]

and there’s something similar for \(N\) odd.

Okay, fine. But then where does the \(-1/12\) come from? Let’s go back to the fake derivation. The \(-3\) factor showed up because of this manipulation:

\[T - 4T = R\]

This is simply not valid if you’re tracking partial sums: \(T_N - 4T_N \? R_N\) is false; the actual relationship is \(T_N - 4T_{\lfloor N/2 \rfloor} = R_N\) But maybe that’s sufficient? After all the constant term of \(R_N\) is going to depend on the constant terms of \(T_N\) and \(T_{\lfloor N/2 \rfloor}\), which depends on \(T_{\lfloor \lfloor N/2 \rfloor/2 \rfloor}\), etc, all the way down to \(T_0 = 0\). So we need to sum over the whole recurrence to see if this is true. It looks like this:

\[T_N = \underbrace{R_N + 4 R_{\lfloor N/2 \rfloor} + 4^2 R_{\lfloor (\lfloor N/2 \rfloor)/2 \rfloor} + \ldots}_{\log_2 N + 1 \text{ terms}}\]

There are \(\log_2 (N) + 1\) terms, since that is the number of times you can repeat this \(N \mapsto \lfloor \frac{N}{2} \rfloor\) calculation (each one being equivalent to lopping off the lowest digit in binary; we would write \(N \gg 1\) if we were in a programming context).

So since \(T_N = R_N + 4T_{\lfloor N/2 \rfloor}\) is defined by a recurrence, there is a series of \(R_N\)s in the resulting sum, and therefore multiple copies of the \(\frac{1}{4}\) constant term from each \(R_N\). Here is an example calculation of \(T_5\) to make it easy to see:

\[\begin{aligned} T_5 &= \boxed{\frac{1}{4}} - (-1)^5 \frac{2(5)+1}{4} + \boxed{4 [\frac{1}{4}} - (-1)^2 \frac{2(2)+1}{4} + \boxed{4 [\frac{1}{4}} - (-1)^1 \frac{2(1)+1}{4}]] \\ &= \frac{1}{4} + (\frac{2(5)+1}{4}) + 1 - 4 \frac{2(2)+1}{4} + 4 + 16 \frac{2(1)+1}{4} \\ &= \frac{1}{4} + \frac{11}{4} + 1 - 5 + 4 + 12 \\ &= 15 \\ &= 1 + 2 + 3 + 4 + 5 \end{aligned}\]

The multiple copies of \(\frac{1}{4}\) form a series that looks like this:

\[\begin{aligned} T_N &\sim \frac{1}{4} + 4 \frac{1}{4} + 4^2 \frac{1}{4} + \ldots \\ &= (\underbrace{1 + 4 + 4^2 + \ldots}_{\approx \, \log_2 N + 1 \text{ terms}}) (\frac{1}{4}) \\ \end{aligned}\]

But we can (divergently) sum that—it’s

\[\begin{aligned} T &\? (1 + 4 + 4^2 + \ldots) (\frac{1}{4}) \\ &\? \frac{1}{1 - 4} (\frac{1}{4}) \\[0.8em] &= -(\frac{1}{3}) (\frac{1}{4}) \\[0.8em] &= -\frac{1}{12} \end{aligned}\]

Evidently this is where the number is coming from. Inside the overall series for \(T_N\) there are a series of \((\log_2 N + 1)\) of these \(4^k \frac{1}{4}\) terms which on their own can be have a constant part of \(-\frac{1}{12}\). They come from the fact that \(T_N\) can be written as a sum of a bunch of \(R_N\), each of which has a constant term. Evidently \(T_N\), with \(N\) terms, ‘contains’ another series inside of it with \((\log_2 N + 1)\) terms, which has constant part \(-\frac{1}{12}\).

The exact value of the whole \((1 + 4 + 4^2 + \ldots) (\frac{1}{4})\) series when we use the exact sum instead of the divergent one is

\[\begin{aligned} T_N &\? (1 + 4 + 4^2 + \ldots) (\frac{1}{4}) \\ &= \frac{1}{4} S_{\log_2 N + 1}(4) \\ &= (\frac{1}{4}) \frac{1 - 4^{\log_2 N + 1}}{1 - 4} \\ &= (\frac{1}{4}) \frac{4N^2-1}{3} \\ &= -\frac{1}{12} + \frac{N^2}{3} \\ \end{aligned}\]

Therefore a complete expression for \(T_N\) is that, plus all the \(\pm\) terms that we dropped:

\[T_N = -\frac{1}{12} + \frac{N^2}{3} + \text{(a bunch of oscillating terms)}\]

And the reason that the \(T - 4T = R\) manipulation appeared to work is that it was actually computing

\[\begin{aligned} T - 4T &= (R_N + 4 R_{\lfloor N/2 \rfloor} + 4^2 R_{\lfloor (\lfloor N/2 \rfloor)/2 \rfloor} + \ldots) - 4(R_N + 4 R_{\lfloor N/2 \rfloor} + \ldots) \\ &= R_N + 4 (\underbrace{\cancel{R_{\lfloor N/2 \rfloor} - R_N}}_{\text{??}}) + 4^2 (\underbrace{\cancel{R_{\lfloor (\lfloor N/2 \rfloor)/2 \rfloor} - R_{\lfloor N/2 \rfloor}}}_{\text{???}} ) + \ldots \\ &\? R_N \end{aligned}\]

Which, again, is definitely invalid in general, but does work if you consider only the constant terms of each of the \(R\) since they are all \(\frac{1}{4}\) with no \(N\)-dependence.

So I guess that’s why the \(-\frac{1}{12}\) shows up: it is really there, but only when you divergently sum up the series created by all \(R_N\)’s constant terms, ignoring their actual value. Including those gives the partial sums the overall shape of a parabola \(-\frac{1}{12} + \frac{N^2}{3}\), and adding back in all the oscillating terms as well gives the full \(T_N = N(N+1)/2\).

I am still not quite sure how to reconcile this expansion with the fact that the closed form for \(T_N\) is \(\frac{N(N+1)}{2}\), though. I guess there’s a sense in which these closed forms want to be written as a sum of oscillating- and non-oscillating terms? Wikipedia suggests that the sense in which it is correct is that, as \(N \ra \infty\), the best parabolic approximation to \(T_N\) is one which ‘smoothes out’ the jumps at each integer discontinuity. For whatever reason constructing \(T_N\) in this way correctly handles that, whereas just writing its value down directly does not.

Okay, that’s enough for now. Perhaps later I will come back and add some more worked-out sums on here. I am fairly confident this sort of technique continues to work on most other divergent sum results, but I’ll have to try them to be sure.

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5. A philosophical note

What I have done with each series here demonstrate something that I have found to be a helpful technique for resolving a class of mathematical paradoxes, in particular anything that involves infinities and limits. The general approach: whenever there is a limit \(\lim_{N \ra a}\) taken in the course of demonstrating some paradoxical outcome, instead of thinking of the value of \(N\) canceling out, think of it as just going unstated in the rest of the problem. Basically think of it as still being there but you’re just electing not to write it anymore. If later on you produce a result which seems paradoxical, go back and restore the \(N\)s throughout the computation and see if they resolve the surprise. Usually they do.

I consider this to be a finitist approach, and the arguments in this paper to me demonstrate the merit of this type of finitism. I keep finding that very many of the things which are otherwise vexing in higher mathematics simply disappear if you insist on finitism: you can’t get surprising results from limits if you refuse to take limits at all! Finitism (or at least, my flavor of it, which I have not seen written down anywhere yet) is so often helpful for ensuring nothing bizarre happens that I have come to think that it will continue to work for anything else I study.

The downside to finitism and the rejection of limits is that much of the content of mainstream mathematics becomes basically unusable—not because it’s wrong or anything, but because it’s just about the wrong stuff. In particular topology and analysis are built entirely on the assumption that numbers can be compared with basically “infinite resolution”, and therefore many of their theorems do not really survive when you reject this assumption. As a result learning these subjects is a slog. On the other hand finitism can maybe save you time when studying physics, because physics is inherently finitistic: you can never have truly infinite quantities (except for self-induced coordinate singularities, which can be handled). Therefore, sticking with finitistic math means that you will never have to worry about mathematical paradoxes sneaking into your physical models which you would then have to deal with. Probably this is beneficial.

The main reason I have written this article is in order to have these calculations set down somewhere so I can not worry about them again. But I also wanted to demonstrate some of the merits of my finitist philosophy. Sticking with finite results whenever possible lets us say only things which are unobjectionably true. We see that for these (comparatively gentle) divergent sums, all of the special ‘techniques’—like analytic continuation, rules about when you can interchange or respace terms, smoothing with mollifiers, whatever else—are not necessary; careful math gives the same answers with less theory and with no question as to the interpretation of the result. I can’t confidently say that this will work in general since I’ve tackled only the easiest examples here, but I do think it is a worthy approach, and far more philosophically sound than the other things I see people trying.