One of my more strongly held mathematical opinions is that results which appear to be surprising or paradoxical (or just overly complex) are in fact bad results, and are often 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 if anything an argument for finitism. Surprising implications of the axiom of choice are reasons to discard the standard forms of set theory as a foundation. That sort of thing. I can’t really justify these claims (yet) but I believe them.

I have long had a vendetta against, in particular, all math explainers who give too much credence to nonsensical divergent series summation results like \(1 + 2 +3 + 4 + \ldots \? -\frac{1}{12}\). I wrote about that sum in one of my first articles, which was inspired by a Numberphile video which had a very disappointing discourse around it at the time. I felt that people were far too willing to say that it was “in some sense” true, or even literally true, because they were basically 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. And if there is one thing that academics should not do it is gaslight the public. Truth is determined by reality, not by fancy techniques inside some formalism we got used to.

Some years on, I have concocted a much more pleasant and simple exposition on divergent series 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 wants to see this laid to rest (or, I guess, thinks there’s something I’m still missing). Nothing in here is particularly deep, I think, and that’s kind of the point. It is just a very simple explanation for a simple thing.

This article is a part of an earlier piece about Taylor series which I decided to factor out and expand into a standalone post. The rest of that post is more of a pile of unhinged musings, whereas this part is quite concrete and more likely to be useful to someone else, so it seemed better to separate the two. I also wanted to add another section that just didn’t belong there. So now it’s its own thing.

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1.

I like to say that

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

is my favorite function, because it has the best Taylor series. Around \(x=0\) it is

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

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

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

We can also write down the Taylor series around \(x=\infty\) by expanding around \(1/x = 0\) instead. It is:

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

Which works for \(\| x \| > 1\). For example,

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

It is interesting to contemplate the fact that plugging \(2\) into the first series sorta works: \(S(2) = 1 + 2 + 4 + 8 + \ldots\) gives a series whose “sum”, by any of the various divergent series summation techniques, equals \(-1\). The standard “proof” 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 is not a number, it still contains the data \(-1\) somehow—maybe in a form like \(-1 + O(\infty)\)? This is an explanation I saw a lot online a while ago. People will often link Tao’s article on the subject which shows how you can think of these sums as showing up more clearly when you smooth out the sum so that does not have discrete jumps at each integer anymore.

But I’ve since come to think that all of those explanations are making things too complicated. There is a very simple way of thinking about this that is completely satisfactory for intuition, which is as follows. Depending on your background it might be a bit elementary, so I feel weird making a big deal about it, but I think it is the only way to stay sane.

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In general it is not the case that \(S(x)\) is the multiplicative inverse of \((1-x)\). This is clear if you write it as a partial sum, \(S(x) = 1 + x + x^2 + \ldots + x^N\), as \(N \ra \infty\). Then

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

So the real value is

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

And for the expansion around infinity, we have \((1-x) S_{\infty}(x) = 1 - x^{-N}\), therefore

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

Normally we think of the value of \(N\) going to infinity, and it is clear why this works in the given radii of convergence: \(S(x) = \frac{1}{1-x}\) if \(\lim_{N \ra \infty} x^{N} = 0\) only. This is why \(S(1/2) = 2\) is correct. Meanwhile

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

is indeed the value of \(1+2+4+\ldots\), but it’s not the value of \(\frac{1}{1-2}\).

The reason that \(S(2) = -1\) under divergent summation techniques is because the algebraic manipulations used are conveniently erasing the \(O(x^{N+1})\) term entirely, even though it should contribute to the result. At the step

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

the algebra just drops whatever happens to the \(O(x^N)\) term even though it is large when \(x=2\). Since it is hiding behind the ellipses, it is easy to ignore. The computation which 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 \(O(2^N)\) term cancelled — you just decided to ignore it. The mistake results from being too used to thinking of \(S(x) = 1/(1-x)\) as being ‘exact’. It never actually was, except arguably in the limit \(N \ra \infty\), but you can’t use that anymore.

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Here is a more complicated example which is resolved by including the \(x^N\) terms:

Consider the function \(g(x) = \frac{1+x}{1+x+x^2}\) which should have \(g(1) = 2/3\). But the summation \(g_0(1)\) appears to give an unsummable divergent series

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

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 value \(g(1) = 2/3\) by l’Hôpital’s rule. But the series expansion in the next step doesn’t work, because the \(O(x^N)\) term is being dropped again. It should be

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

This time the remainder term that is normally 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.

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I tend to be more of a finitist than is commonly acceptable. Any mathematician will tell you that the above partial sums are true, but might note that they skirt the question of taking the limit as \(N \ra \infty\), which is what makes \(S(1/2) = 2\). Indeed, what makes divergent sums mysterious is how they seem to work after taking limits, despite the limits being invalid.

Personally I don’t buy this. I think 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.

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2

\(S(x) = 1+x+x^2 +\ldots\) is admittedly the easiest of the divergent series to figure out what’s going on in. Therefore, to prove that this way of thinking works more generally, and also mostly just to have a record of them, I will go through each of the other interesting divergent sums I know and show how this ‘finitistic’ perspective gives the right answer.

The first series to consider is

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

A common 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.

Clearly if we fix \(P\) to have a finite number of terms \(N\) then its sum is actually \(P = N\) (this is also the value of \(S(1)\)—clearly not equal to \(1/(1-1)\) by the way!). To handle this better, from now on we will label series with a subscript to indicate by how many terms are included in them, like this:²

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

Then we can say concretely:

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

This always works and is basically trivial.

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Next we have

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

Which is \(S(-1)\), summable with the classic manipulation

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

The actual value is clearly

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

So we repeat the calculation:

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

turns out 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 this sum: it has an \(N\)-dependent term at the end whose value is \(2/3\). No sums of integers had to equal fractions at all. Whew. (It will not be surprising that you can also get a version of \(1-1+1-1+\ldots\) that equals any rational you want, by computing \((1-x^m)/(1-x^n)\) instead.)

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Next up,

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

with the classic manipulation

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

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? After some thought: the partial sums of \(R\) are

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

which are a lot like \((1, \frac{3}{2}, 2, \frac{5}{2}, \ldots)\), except that every other term is wrong. We can write it like this, though:

\[(1, \frac{1}{4} - \frac{5}{4}, 2, \frac{1}{4} - \frac{9}{4}, 3, \frac{1}{4} - \frac{13}{4}, \ldots)\]

and so on. That is,

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

Therefore the closed form must be

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

Which is why the invalid computation that canceled out the \(N\)-dependency gave \(R = 1/4\).

Another hacky way of getting \(R=1/4\) is to try squaring \(Q\), and then arranging the terms in a certain way:

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

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 \(R\). If we want to use this to compute any particular value of \(R_N\), we need the sum to instead be

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

such that every sum terminates on exactly the \(N\)th term of \(1-2+3-4 +\ldots\). This is not the same as \(Q_N^2\). Since \(Q_N\) is \(0\) for \(N\) even and \(1\) for \(N\) odd, this simplifies to

\[\begin{aligned} \underbrace{Q_N - Q_{N-1} + Q_{N-2}-Q_{N-3} + \ldots \mp Q_0}_{N \text{ terms}} &= \begin{cases} -\frac{N}{2} & N \text{ even} \\[0.5em] \frac{N+1}{2} & N \text{ odd} \end{cases} \\ &= R_N \end{aligned}\]

So that’s good. But it’s not what you get from computing \(Q_N^2\), because the \(N\)-dependent terms are all wrong. Still, the constant term is valid, which is why \(Q^2 = \frac{1}{4} = R\) seems to work.

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Next up, the Ramunajan sum

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

Which can be ‘summed’ by reducing to \(R\):

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

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

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

which decidedly does not contain a \(-\frac{1}{12}\) term. So where did 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.

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

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 \(T_N = N(N+1)/2\) obeys this relation. For example if \(N\) is even then \(N(N+1)/2 = \frac{N^2}{2} + \frac{N}{2} = -\frac{N}{2} + (N + \frac{N^2}{2}) = -\frac{N}{2} + 4(\frac{(N/2)(N/2+1)}{2})\).

Okay, fine. But then where does the \(-1/12\) come from?

Back to the fake derivation. The \(-3\) factor shows up because of this manipulation:

\[T - 4T = R\]

Which is simply not valid if you’re tracking partial sums, since \(T_N - 4T_N \? R_N\) is false; the actual relationship is \(T_N - 4T_{\lfloor N/2 \rfloor} = R_N\). What we can do, though, is track the remainder and see what it does. We get

\[T_N - 4 T_N = -3 T_N = R_N - 4 [T_N - T_{\lfloor N/2 \rfloor } ]\]

Expanding:

\[\begin{aligned} T_N &= -\frac{1}{3} R_N + \frac{4}{3} [T_N - T_{\lfloor N/2 \rfloor } ] \\ &= -\frac{1}{12} + \frac{1}{3} [(-1)^N \frac{2N+1}{4}] + \frac{4}{3} [\, \boxed{\frac{1}{4}} - (-1)^N \frac{2N+1}{4} + 4 T_{\lfloor N/2 \rfloor} - T_{\lfloor N/2 \rfloor }] \\ &= \frac{1}{4} - (-1)^N \frac{2N+1}{4} + 4 T_{\lfloor N/2 \rfloor} \end{aligned}\]

We see that there is another constant term which modifies the leading \(-\frac{1}{12}\). It come from the \(T_N\) in the remainder, which was expanded as \(T_N = R_N + 4T_{\lfloor N/2 \rfloor}\) in the first place. By ignoring this remainder we end up with a ‘dangling’ \(-\frac{1}{12}\) which appears to be the constant term in the whole sum.

When we include it we get a \(+\frac{1}{4}\) constant term instead, which is still not the \(0\) we expect from the formula \(\sum^N_1 k = \frac{N(N+1)}{2}\), however. I suppose it must come from including the \(T_{\lfloor N/2 \rfloor}\) terms as well? For instance if \(N=5\) and we write the whole sum out,

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

I guess you can’t tell anything about the asymptotic behavior from this, but it is at least interesting to see how it works. There are a bunch of constant terms, \(1 + \log_2 (N)\) of them (so only one is actually ‘constant’), plus a bunch of ‘oscillating’ terms with different magnitudes which serve to create all the cancellation necessary to make the result.

So on the one hand we know that \(\sum_1^N k = \frac{N (N+1)}{2}\) and there is no constant term, and we have demonstrated that correctly doing algebra with \(T_N\) does not produce a \(-1/12\), and incorrectly doing algebra produces one because the term that should cancel it out is being swept under the rug as a missing remainder in \(T - 4T = R \;\; (+ [-4 T_N - T_{\lfloor N/2 \rfloor }])\). And yet—Tao’s mollifier arguments, and all the graphs on Wikipedia that show that a parabola approximation to the discrete sum goes through \(y=-1/12\), must also be correct. And then there’s those people who claim this sum shows up in quantum field theory and the \(-1/12\) means something. How do we reconcile these facts?

I have one idea. Consider what happens to the \(T_N = R_N + 4 T_{\lfloor N/2 \rfloor}\) recurrence as \(N\) gets large:

\[T_N = R_N + 4 R_{\lfloor N/2 \rfloor} + 4^2 R_{\lfloor (\lfloor N/2 \rfloor)/2 \rfloor} + \ldots\]

There are \(\log_2 N + 1\) of these terms (I think?), and every one of them contributes a copy of \(\frac{1}{4}\), meaning that the resulting ‘constant’ part of the sum (which is not actually constant since there are a varying number of them) is

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

If we then apply divergent series summation trickery to this sum, we get what we’re looking for:

\[(1 + 4 + 4^2 + \ldots) (\frac{1}{4}) \? \frac{1}{1 - 4} (\frac{1}{4}) = -(\frac{1}{3}) (\frac{1}{4}) = -\frac{1}{12}\]

So I think that may be where the number is actually 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 divergently-summed to \(-\frac{1}{12}\). The source of them is basically from the fact that \(T^N\) can be written as a sum of a bunch of \(R_N\), each of which does have a constant term. Perhaps a way of saying this is that: \(T_N\) does not have a constant term like \(R_N\) does for finite \(N\), but as \(N \ra \infty\) it does split into a constant term and a term which is proportional to \(4^{\log_2 N} = N^2\). Indeed, the actual sum of those terms is

\[(\frac{1 - 4^{\log_2 N + 1}}{1-4})(\frac{1}{4}) = \frac{4N^2-1}{3} (\frac{1}{4}) = -\frac{1}{12} + \frac{N^2}{3}\]

Giving the overall form of \(T_N\) as

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

if I did not screw up my arithmetic anywhere. So I guess that’s why the \(-\frac{1}{12}\) shows up: it is really there, but really only when you consider the series for large \(N\), whereupon the overall shape of the partial sums starts to force this dominant term to appear in any e.g. parabolic approximation. I don’t know enough to say that this is rigorous, but I think it’s right.

Amusingly, the reason the \(T - 4T = R\) manipulation worked is that it was sort of actually computing

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

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

I still am not quite sure how to reconcile this expansion with the fact that \(T_N = \frac{N(N+1)}{2}\) though. It seems like the \(-\frac{1}{12}\) version is somehow talking about a constant term specifically with respect to the \((-1)^N\) oscillations that are introduced by the \(R_N\)s, which \(\frac{N(N+1)}{2}\) glosses over and therefore does not detect. But I don’t know how to think about it really.

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

The point of all this is mostly to demonstrate my philosophy of staying finitist whenever possible. At least for these (comparatively gentle) divergent sums, special ‘techniques’ like analytic continuation, rules about when you can interchange or respace terms, or smoothing with mollifiers are not necessary: careful math gives the same answers, with less theory, and with no question as to the interpretation of the result. I’d like to say I’m confident that this will hold in general, but of course I’m not. But I do think it is a worthy approach, and far more philosophically sound than the other things I see people trying.