The repair at the boundary

The yardstick ends on a gap, and the gap is exactly where the subject’s boundary is. Every series $1/n^p$ has the ratio of consecutive terms tending to one, so the ratio test decides none of them, and the family contains both the standard convergent series and the standard divergent one.

The repair is a move already used once, on one series, without its generality being noticed.

That is Cauchy condensation, and the whole of it is in the caption. Replace every term of a run by the biggest term in it, and a series with no ratio becomes a series with one.

The move, and where it came from #

The construction is Oresme’s, from the fourteenth century, and it has already appeared once here to show the harmonic series passes every bound: group its terms into runs of 1, 2, 4, 8 terms, note that each run’s terms are all at least the last one in it, and conclude that each run totals at least $2^k \cdot 2^{-(k+1)} = 1/2$. Infinitely many halves pass every bound.

The observation that makes it an instrument rather than a trick is that nothing in that argument used the terms being $1/n$. Take any series whose terms are positive and decreasing. Bracket them into runs of doubling length. Then for every run:

• every term is at most the run’s first term, so the run totals at most $2^k a_{2^k}$;
• every term is at least the run’s last term, so the run totals at least $2^{k}a_{2^{k+1}}$, which is half of $2^{k+1}a_{2^{k+1}}$.

So the original series and the condensed series $\sum 2^k a_{2^k}$ are within a factor of two of each other, in both directions, at every stage. A factor of two cannot turn a finite total into an infinite one. The two series converge together and diverge together, and the condensed one is often trivial.

The hypothesis that the terms decrease is not decoration; it is what lets a run be bounded by one of its own members. A series whose terms wander is not condensed by this argument, and no repair makes it so.

The threshold, in one line #

Apply it to $a_n = 1/n^p$. The condensed terms are

$$2^k \cdot \left(2^k\right)^{-p} = \left(2^{1-p}\right)^k,$$

which is a geometric series with ratio $2^{1-p}$. That ratio is below one exactly when $p > 1$.

So $\sum 1/n^p$ converges exactly when $p$ exceeds one, and the whole family is decided by one exponent comparison. No integral, no limit comparison, no case analysis — the boundary is at the place where $2^{1-p}$ crosses one, and it crosses at $p = 1$ because that is the exponent at which doubling the run length exactly cancels the terms getting smaller.

That last sentence is the content, and it is worth saying without the algebra. A run of doubling length has twice as many terms as the one before. Whether its total grows or shrinks is a race between the count doubling and the terms shrinking. At $p = 1$ the terms halve over a run and the count doubles, and the two cancel exactly — every run totals about the same, so the total climbs for ever by a fixed amount each time. Above $p = 1$ the terms halve faster than the count doubles and the runs shrink geometrically.

Three drawings of the same bracketing at three exponents, and the ratio of the outlined bars is the entire verdict in each. The picture is the same picture; only the direction the outlines go changes.

Watching the threshold at scale #

The plot is worth reading carefully, because it shows what the threshold looks like to somebody adding the terms up rather than proving things about them — and the answer is: almost nothing.

At $p = 0.9$ the series diverges and after a hundred thousand terms its partial sum is around twenty-two. At $p = 1.1$ the series converges to about $10.6$ and after a hundred thousand terms it has reached around $9.5$. The two curves are close together, both still climbing, and nothing a computation of any feasible length could distinguish separates them. The classification is exact and the observation is hopeless, which is the standing relationship between these two ways of asking about a series.

The straight line at $p = 1$ is the one feature the plot does show honestly. A partial sum that is a straight line against $\log n$ is a partial sum that is a logarithm, and a logarithm passes every bound — slowly enough that a million terms reach only fourteen, which is why the divergence was not obvious to anybody before Oresme and was doubted for centuries afterwards.

Doing it again #

The threshold is at $p = 1$ and $p = 1$ is not in the convergent family, so the obvious next question is what sits immediately above the harmonic series. Nothing of the form $1/n^{1+\varepsilon}$ is immediately above it — halving $\varepsilon$ gives something between. The right place to look is at a series that is $1/n$ multiplied by something that shrinks more slowly than any power.

Condense $a_n = 1/(n\,(\log n)^q)$. The condensed term is

$$2^k \cdot \frac{1}{2^k (\log 2^k)^q} = \frac{1}{(k \log 2)^q},$$

which is $1/k^q$ up to a constant. So the whole question has moved down one level, and the threshold for the new family is at $q = 1$ by the result of the last section.

So $\sum 1/(n\log n)$ diverges and $\sum 1/(n(\log n)^2)$ converges, and both lie strictly between $\sum 1/n$ and every convergent $p$-series. That is already enough to say something about the structure:

Between the divergent series and the convergent ones there is no gap to stand in.

And the construction does not stop. Condense $1/(n \log n (\log\log n)^q)$ and the same computation puts the threshold at $q = 1$ again, one level further down. Repeat for ever. There is a series between any two adjacent members of the scale, in both directions, at every level — this is the Abel–Dini hierarchy, and it has a theorem attached that is stronger than any example:

Given any divergent series of positive terms, a divergent series with strictly smaller terms can be constructed from it; and given any convergent one, a convergent series with strictly larger terms. The construction is short and worth having: for a divergent $\sum a_n$ with partial sums $s_n$, the series $\sum a_n / s_n$ also diverges and its terms are strictly smaller from the moment $s_n$ exceeds one. Applying it to the harmonic series gives $\sum 1/(n H_n)$, and $H_n$ behaves like $\log n$ — which is the same logarithm the collector’s wait is made of, arriving here as the thing that makes a divergent series diverge more slowly. So there is no slowest divergent series and no fastest convergent one, and no single comparison decides everything.

What that costs, and what it does not #

The hierarchy has a consequence that is often stated too strongly and is worth stating exactly.

There is no universal comparison series. Any fixed convergent series fails to dominate some convergent series; any fixed divergent series is not dominated by some divergent one. So no single instrument of the comparison kind can decide every case, and the search for one is not merely unfinished but impossible.

What is not true is that there is no general test. Condensation itself is general, within its hypothesis: give it any positive decreasing terms and it returns an equivalent series, and repeating it walks down the hierarchy as far as anybody needs. Kummer’s test is general in a stronger sense — for any positive series there exists a choice of auxiliary sequence that decides it — which is true and nearly useless, since finding the sequence is the original problem.

The honest summary is that the boundary is not a place but a direction, and each of these tests names a rate that a series is compared against. There are always rates in between; the tests get finer indefinitely; and no finite list of them is complete.

Two places the threshold turns up already #

The exponent one is not an isolated fact about a family of series, and two of the places it has already appeared in this collection are worth setting beside it, because they are the same threshold arrived at from directions with no series in them.

The first is about lattice points. Counting how many whole-number points lie inside a circle of radius $R$ gives the area plus an error, and how large that error can be is a question about how much cancellation there is in a sum of arithmetic terms. The exponents argued over there — a half against two thirds — are the exponents of exactly this kind of comparison, and the reason the question is hard is that the series involved sit near their own boundary.

The second is about how closely a fraction can approximate. The measure-theoretic statement that almost every number is badly approximable past a certain exponent is proved by a covering argument whose total length is a $p$-series, and the threshold at $p = 1$ is what decides it — the same covering a set of measure zero is built from, where a family of intervals of total length under any bound one likes is what makes a set of measure zero.

Neither of those is a series question on its face. Both reduce to one, and both reduce to a $p$-series, and in both the answer hangs on the same exponent. That is what it means for a threshold to be structural rather than an artefact of a family: it is where a count doubling and a quantity halving cancel, and that race turns up wherever something is being summed over scales.

The same argument in its other clothes #

The integral test decides the same family, gives the same threshold, and is the same argument with the runs made infinitesimal.

For positive decreasing $f$, the terms $f(n)$ bracket the area under $f$ between consecutive integers from both sides, so $\sum f(n)$ and $\int_1^\infty f$ converge together. For $f(x) = x^{-p}$ the integral is elementary, it is finite exactly when $p>1$, and the whole question is over.

The two arguments bound the same sum by the same manoeuvre at different granularities. Condensation replaces a run of terms by its largest and smallest; the integral test replaces one term by a strip. Condensation’s runs grow, which is what turns a $p$-series into a geometric one; the integral’s strips do not, which is what makes it an antiderivative problem instead. Whichever is easier depends entirely on whether the function has an antiderivative, and for $1/(n\log n\log\log n)$ it does — so the integral test walks down the hierarchy just as far, one substitution at a time.

What the integral test adds is the rate, which condensation does not give. The tail of $\sum 1/n^2$ past $N$ is between $1/(N+1)$ and $1/N$, read straight off the two integrals, and that is the estimate the comparison against a geometric series could not produce.

How slowly the boundary is crossed #

The classification is exact and the arithmetic either side of it is unreasonable, and it is worth putting numbers on that, because the two facts together are what makes the threshold a piece of mathematics rather than a piece of computation.

The harmonic series reaches $10$ after about $12{,}367$ terms, $20$ after about $272$ million, and $100$ after roughly $1.5 \times 10^{43}$ — the count needed multiplies by about $e^{1}$ for each extra unit, so every step costs $2.7$ times the terms the last one did. Adding it up to pass $100$ is not slow, it is impossible; and the series nonetheless passes every bound, which is a statement no computation will ever witness.

Just above the threshold the situation is the mirror image. $\sum 1/n^{1.01}$ converges to about $100$, and its partial sum after a million terms is around $87$ — still thirteen short, with the remaining thirteen spread over the next $10^{200}$ terms. A computation that stopped at a million and reported $87$ would be reporting a number that is neither the limit nor close to it, and nothing in the arithmetic would say so.

So both sides of the boundary are computationally indistinguishable from each other and from their own answers. That is the argument for the proof: a verdict that cannot be reached by adding is a verdict that has to come from an inequality, and the inequality here is the one the bracketing produces in a line.

It is also why the divergence of the harmonic series stayed contested for so long after Oresme. His argument is complete and was published around 1350; it was lost, rediscovered by Mengoli in 1650 and by the Bernoullis in the 1680s, and in between there was no shortage of people who had added the terms up and found them settling. They were right about what they saw, which is the whole difficulty.

A boundary with nothing at it cannot be drawn #

The runs are drawn to seven and the claim is about all of them. Each figure’s bars are a finite prefix of an infinite bracketing, and the check made where the figure is computed — that every drawn run sits between its two bounds, and that the upper bound is exactly the $k$-th power of $2^{1-p}$ — is a check on the arithmetic at those runs and not a proof of the inequality.

The threshold cannot be drawn at all. No figure here shows $p = 1$ being the boundary; the three condensation figures show three exponents, one on each side and one on it, and the statement that nothing between $1$ and $1+\varepsilon$ changes the verdict is a consequence of the formula rather than of the picture. A drawing at $p = 1.001$ would be indistinguishable from the drawing at $p = 1$.

And the hierarchy is drawn one level deep. The log-log figure shows the second level; the third level differs from it by a factor that has no visible effect at a million terms, and the claim that the construction never terminates is an induction with no picture anywhere in it.

The partial-sum plots are the most misleading figures in the essay and are included for that reason. They show two curves nobody could separate belonging to series with opposite answers, which is the true state of affairs and looks like an inadequacy of the drawing. It is not: no amount of computation separates them either.

Still open here: a boundary with no members in it #

Condensation settles the $p$-series and opens a question it cannot close. The scale of rates between convergence and divergence is dense, so the boundary is a cut with nothing at it — and a comparison test is an instrument that needs something to compare against.

Two directions lead out. One asks what can be said about a series’ rate without naming a comparison at all, which is where the summability methods of the rearrangement essay start: assign a value by averaging rather than by adding, and some divergent series acquire one — which is also what rescues the formula that keeps going past its own interval at the boundary. The other asks what the hierarchy looks like as an object — it is an ordered family with no gaps and no ends, which is the shape of a question about orderings rather than about series, and the answer is that no countable list of rates is cofinal in it.

One move, used three times #

The habit is the thing to keep, and it is smaller than the theorem.

The harmonic series’ divergence, the $p$-series threshold and the whole hierarchy above it are one manoeuvre: replace a run of terms by copies of the largest, and choose the run lengths so that the replacement is geometric. Oresme used it on one series and it looked like a trick for that series. It is a trick for every decreasing series, and the run lengths doubling is the only choice in it.

The general shape is worth recognising elsewhere. When an instrument cannot see a distinction, the useful move is often not a finer instrument but a regrouping of what is being measured — so that the quantity the instrument does see becomes the quantity in question. Here the instrument reads ratios, the series has none, and the regrouping manufactures one.