I tend to think that results which appear to be surprising or paradoxical (or even just overly complex) are in fact bad results, and are better thought of as indicators of places that our foundations have gone wrong. To me 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 a finistic theory which excludes them, and surprising implications of the axiom of choice are evidence that we should discard set theory as a mathematical foundation rather than interesting discoveries. That sort of thing. I can’t really argue convincingly for these claims (yet) but I believe them.

In particular I have a vendetta against all the online math explainers who have given too much credence to nonsensical manipulations of infinite quantities—in particular, to those which lead to identities like the infamous \(1 + 2 + 3 + 4 + \ldots \? -\frac{1}{12}\). I wrote about that sum in one of my first articles, which was inspired by a Numberphile video that had a very disappointing discourse around it at the time. People were far too willing to say that the sum equaling \(-1/12\) was “in some sense” true, or even literally true, basically because they were unable to bring themselves to say that their mathematical understanding had a hole in it—even though any layperson could clearly see that it did. If there is one thing that academics should not do it is gaslight the public. If your mathematical framework disagrees with reality then it’s a bad framework and you should stop using it.

I have since found a much more simple and pleasing exposition on these sorts of sums than I had before, 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.

(The first 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 added some more material in the process.)

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1. Geometric Series

First we will discuss the best Taylor series, which is that of

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

Around \(x=0\) it is

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

This converges for \(\| x \| < 1\). For example,

\[\frac{1}{1-\frac{1}{2}} = 1 + \frac{1}{2} + \frac{1}{4} + \ldots = 2\]

This equality holds if the \(1+1/2+1/4+\ldots\) is regarded as a limit, which is standard. It’s equivalent to how \(1.99\overline{9} = 2\), and 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.

We can also write down a Taylor series for \(f(x)\) around \(x=\infty\): it’s the same series except you write everything in terms of \(1/x\) first. I’ll write this as \(S_{\infty}(x)\) to distinguish it from \(S(x)\):

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

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

\[\frac{1}{1-2} = -\frac{1}{2} \frac{1}{1 - \frac{1}{2}} = -\frac{1}{2}[1 + \frac{1}{2} + \frac{1}{4} + \ldots] = -1\]

There are also series around any \(k \neq 1\), but we don’t need those.

Outside of the radii of convergence of these series they give infinite results which are not equal to \(\frac{1}{1-x}\) at the same point. For instance \(S(x)\) converges for \(\| S \| \leq 1\), and outside of that it diverges to infinity:

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

However, it is interesting to contemplate the fact that this series does seem to encode the value of \(f(2) = 1/(1-2) = -1\) somehow. It’s a divergent series whose sum, if evaluated by any of the various divergent series summation techniques, somehow equals \(-1\) anyway. A standard “proof” of this fact is

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

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 \(-1 + O(\infty)\)? This is an explanation I have seen online a few times. People will often 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.

However, I’ve since come to think that all of that is making things too complicated. There is a very simple way of thinking about this that is completely satisfactory for intuition, which I will describe next. 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 and only allow ourselves to regard them as the limits of explicit partial sums.

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2. Exact Geometric Sums

Recall that it is not the case that \(S(x) = 1 + x + x^2 + \ldots\) is generally equal to \(f(x) = \frac{1}{1-x}\), the multiplicative inverse of \((1-x)\). This is clear if you write it as an arbitrary partial sum, \(S(x) = 1 + x + x^2 + \ldots + x^N\). Then, were it the inverse, we would have \((1-x) S(x) = 1\), but we don’t. Instead we have

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

Evidently the real value \(S(x)\) is

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

Likewise \(S_{\infty}(S)\) is not the inverse of \((1-x\), because \((1-x) S_{\infty}(x) = 1 - x^{-N}\). It is really

\[S_{\infty}(x) = \frac{1}{1-x} - \frac{x^{-N}}{1-x} = \frac{1-x^{-N}}{1-x}\]

Normally we think of the value of \(N\) as implicitly going to infinity. It is clear why this leads to each approximation having its corresponding radius of convergence: \(S(x) = \frac{1}{1-x}\) is true only if \(\lim_{N \ra \infty} x^{N} = 0\), and \(S_{\infty}(x) = \frac{1}{1-x}\) only if \(\lim_{N \ra \infty} x^{-N} = 0\). 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 \(S(2) = -1\) diverges is that

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

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 \(S(2) \? -1\) 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, in the version I showed above, it happens at the step where we subtracted the series from itself. It should have read:

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

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 thought we had gotten away it. In fact the computation that “gave” \(S(2)=-1\) should have read

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

which is correct for all \(N\).

When you write \(S(2) = -1\), it’s not that the \(2^{N+1}\) term cancelled — it’s just that you assumed it was zero and dropped it, aided by the fact that writing it as \(1+x+x^2+x^3+\ldots\) notation made it easy to do so. The mistake results from being too used to thinking of the \(x^N \ra 0\) approximation as being exact when it is not. But there is always an “ambient” \(N\) variable which parameterizes how inexact it is.

The same argument gives \(f(-2) = 1 - 2 + 4 - 8 + \ldots\) as

\[S(-2) = 1 - 2 + 4 - 8 + \ldots = \frac{1}{1-(-2)} + \frac{(-2)^{N+1}}{1-(-2)} = \frac{1}{3} + \frac{1}{3} (-2)^{N+1}\]

which is why divergent sum techniques that drop the \(N\) dependency give it the value \(\frac{1}{3}\).

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Apparently it is the case that \(S(2) = 1 +2 +4 + \ldots\) contains a \(-1\) term, since it equals \(2^{N+1} - 1\). And evidently the computation that we’re doing is somehow erasing the \(N\)-dependent part and extract the \(-1\) on its own. I suppose one might interpret the list of divergent series summation techniques as a list of tricks for doing this. This doesn’t make them correct—they’d be better explained as that, rather than as some mysterious way of assigning a true value to the sums. But just in case it 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 should have \(g(1) = 2/3\). The summation \(g(1)\) appears to give an unsummable divergent series

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

The manipulation \(\frac{1+x}{1+x+x^2} = \frac{1-x^2}{1-x^3}\) is valid, and you can tell that it really does have the value \(g(1) = 2/3\) in the limit \(x \ra 1\) by l’Hôpital’s rule. But there’s no way to get \(\frac{2}{3}\) from the series \(1-1+1-1\ldots\).

Things make more sense if we include 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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3. Working up to -1/12

\(S(x) = 1+x+x^2 +\ldots\) is, admittedly, the easiest divergent series for figuring out what’s going on. 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.)

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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.

Although any mathematician will tell you that the above partial sums are correct (assuming I made no algebra errors, I mean), they may object that I have just skirted around the real issue, which is of taking the limit as \(N \ra \infty\). After all that is what \(S(1/2) = 2\). Indeed, what makes divergent sums mysterious is how they seem to work after taking limits, despite the limits being invalid.

I don’t buy this objection. I think the limits don’t really 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).

Analysis is developed by insisting that this hold for all \(\e\), but physics (and all practical applications) doesn’t care; you just need it to be true for the \(\e\) you actually have, which will necessarily be finite. Therefore one does not need to really consider limits, except insofar as they summarize this approximation process succinctly. If they do not, you pull your \(N\)-dependencies back out—hopefully you did not forget where they are—and keep going. True, this is not what is normally meant by the use of limits in math, but my stance is that it should be.

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.

There is a downside, however. So much of mainstream mathematical theory is almost unusable from a finitistic perspective that actually studying anything becomes a slog. Topology and analysis, in particular, are essentially built on the assumption that numbers can be viewed with “infinite resolution”, and many of their theorems do not readily carry over into the setting where you don’t use this assumption anymore. As a result I have to carefully translate their material into finistic terms myself in order to actually do anything to my satisfaction, and this makes absorbing new subjects very slow.

On the other hand finitism can maybe save you time when studying physics. It seems to me that physics is inherently finitistic: you can never have truly infinite quantities except for self-induced coordinate singularities (which can be repaired). 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.