No integrand sits on the border

An endless region with a finite area found the threshold for powers. The region under $1/x^p$ from $1$ to infinity has finite area exactly when $p > 1$: under $1/x^{1.001}$ it is $1000$, under $1/x$ it is infinite. For powers of $x$, the border is sharp and it sits at $p = 1$.

It is tempting to conclude that $1/x$ is the border — that an integrand smaller than $1/x$ eventually has a finite integral and one larger does not. The first half is false. The integrand $1/(x \ln x)$ is smaller than $1/x$ by a factor that grows without bound, and its integral is still infinite. And there is nothing special about that example: below every divergent integrand there is another that still diverges, and above every convergent one there is another that still converges, and the two families press towards each other without ever meeting.

The figure multiplies every integrand by $x$ so that $1/x$ becomes the flat line at height $1$, and plots against the number of digits of $x$, out to numbers with three hundred digits. The solid curves are $1/x$, $1/(x \ln x)$ and $1/(x \ln x \ln\ln x)$: every one of them has infinite area under it. The dashed curves are $1/(x (\ln x)^2)$ and $1/(x \ln x (\ln\ln x)^2)$: every one of them has finite area. Each solid curve lies below the one before it; each dashed curve lies above the one before it. The border between them is not a curve at all.

Dividing by the running integral #

The mechanism that produces the divergent chain is a single step, and it works on any divergent integrand whatsoever.

Let $f$ be positive with $\int^\infty f = \infty$, and let $F(x) = \int_a^x f$ be its running integral, which grows without bound. Consider the new integrand $f/F$. It is smaller than $f$ by a factor $F$ that goes to infinity. Its integral is

$$\int_a^x \frac{f(t)}{F(t)}\,dt = \ln F(x) - \ln F(a),$$

by the substitution $u = F(t)$ — and since $F$ grows without bound, so does $\ln F$. The new integrand still diverges.

Start from $1/x$, whose running integral is $\ln x$: the step gives $1/(x \ln x)$, with running integral $\ln \ln x$. Apply the step again: $1/(x \ln x \ln\ln x)$, with running integral $\ln\ln\ln x$. Each new integrand is smaller than the last by a factor tending to infinity, and each still diverges, as the figure shows — the three running integrals at a number with three hundred digits are about $686$, $5.5$ and $1.9$.

The divergence is real and it is almost invisible. The third integral would reach $10$ only when $\ln\ln x = e^{10} \approx 22{,}026$, that is, when $x$ has about $10^{9566}$ digits. No computation, no physical process and no count of anything in the universe will ever see it pass $10$. It passes every bound nonetheless.

The discrete form of this step is due to Niels Henrik Abel and Ulisse Dini: if $\sum a_n$ diverges with partial sums $S_n$, then $\sum a_n / S_n$ diverges too, while $\sum a_n/S_n^{1+\varepsilon}$ converges for every positive $\varepsilon$. Applied to a sum whose terms vanish and whose total does not — the harmonic series, with $S_n \approx \ln n$ — it produces $\sum 1/(n \ln n)$, divergent, and $\sum 1/(n (\ln n)^2)$, convergent.

The step is the chain rule run backwards #

The computation that makes the step work deserves a second look, because it is the whole argument. The derivative of $\ln F(x)$ is $F'(x)/F(x)$, by the chain rule, and $F' = f$ because $F$ is the running integral of $f$ — the fundamental theorem that area is the undoing of slope established. So $f/F$ is the derivative of $\ln F$, and its integral is $\ln F$ up to a constant.

Nothing about $f$ was used except that it is positive and that $F$ grows without bound. That is why the step applies to every divergent integrand, not only to the logarithmic chain: $f$ could be $1/\sqrt x$, or $e^{\sin x}$, or a step function that is $1$ on scattered intervals and $0$ between them, and $f/F$ would still diverge, more slowly. The logarithm is not a choice made to produce slow functions; it is what the chain rule returns when a function is divided by its own antiderivative.

The same reading explains why the power $1 + \varepsilon$ restores convergence. The integral of $f/F^{1+\varepsilon}$ is $-F^{-\varepsilon}/\varepsilon$ up to a constant, and $F^{-\varepsilon}$ tends to nought, so the integral is finite — for every positive $\varepsilon$, however small. The divergent step and the convergent step sit on either side of the exponent $1$, exactly as the powers of $x$ did, one level down.

The same trick on the convergent side #

On the other side of the border the step is different, and it was found later, by Jacques Hadamard in 1894.

Let $g$ be positive with a finite integral to infinity, and let $R(x) = \int_x^\infty g$ be its tail, the area still to come, which shrinks to nothing. Consider $g/\sqrt R$. It is larger than $g$ by a factor that goes to infinity, since $R \to 0$. Its integral is

$$\int_a^\infty \frac{g(t)}{\sqrt{R(t)}}\,dt = 2\sqrt{R(a)},$$

by the substitution $u = R(t)$ — finite. So the new integrand still converges, and decays more slowly than the old.

Start from $1/x^2$, whose tail is $1/x$: the step gives $1/x^{3/2}$, whose tail is $2/\sqrt x$, and the next step gives a constant times $1/x^{5/4}$. Each step halves the distance of the exponent from $1$. The exponents creep down towards the divergent value $1$ — $2$, $1.5$, $1.25$, $1.125$, $1.0625$ — and never reach it. Once the exponents are close to $1$ the same step, applied to $1/(x(\ln x)^2)$, produces the logarithmic convergent chain in the first figure.

Neither family has an extreme member. There is no slowest divergent integrand, because dividing by the running integral makes a slower one; there is no slowest convergent integrand, because dividing by the square root of the tail makes a slower one.

Slower than a whole chain at once #

That still leaves room for a border of a subtler kind. Perhaps some single integrand $h$ lies below every divergent integrand in a whole chain and above every convergent one — not eventually equal to any of them, but separating them all. Paul du Bois-Reymond proved in 1873 that even this is impossible, and his argument is a diagonal construction.

Take any sequence of divergent integrands $f_1, f_2, f_3, \dots$, each eventually smaller than the one before. Build a new integrand $g$ in blocks. On the first block, let $g = f_1$, and make the block just long enough that the area under $f_1$ over it is $1$ — possible, because $f_1$ diverges. On the second block, switch to $f_2$ and continue until its area over the block is $1$. And so on, forever.

The area under $g$ is $1 + 1 + 1 + \cdots$, infinite. But beyond the start of the $k$-th block, $g$ is equal to $f_k$ or to a later, smaller member of the chain, so $g$ is eventually at most $f_k$ — for every $k$. So $g$ diverges and is eventually below every member of the chain. No chain of divergent integrands is a lower barrier: there is always a divergent integrand below it. The same construction with tails in place of areas works on the convergent side, and combining the two shows that no integrand can lie between the families, whichever chains are chosen.

The blocks in the figure show why this is only a picture of the construction. The first block ends at $x \approx 41$. The second, using $1/(x \ln x)$, needs its logarithm of a logarithm to grow by $1$, and ends at a number with about four and a half digits. The third needs its triple logarithm to grow by $1$ and ends at a number with about $234$ digits. The fourth block would end at a number whose number of digits has more than two hundred digits.

Why the order is only eventual #

Every comparison in this essay is eventual: $1/(x \ln x)$ is smaller than $1/x$ only once $\ln x > 1$, that is for $x > e$, and $1/(x \ln x \ln \ln x)$ is smaller than $1/(x \ln x)$ only once $\ln\ln x > 1$, for $x > e^e \approx 15.2$. Below those points the order reverses, and near the starting point each new integrand is enormous — the logarithms in its denominator are small or negative there.

This is not a defect of the examples. Convergence of an integral to infinity is decided entirely by the integrand’s behaviour far out: changing $f$ on any bounded interval changes the area by a finite amount and cannot turn a finite area into an infinite one. So the only comparisons that can bear on convergence are the eventual ones, and every statement about the border — that one integrand lies below another, that a chain is decreasing — is a statement about all sufficiently large $x$, with “sufficiently large” allowed to depend on the pair. The figures begin their horizontal axes at numbers with a few digits precisely so that the eventual order has taken hold.

The same eventual comparison is what makes du Bois-Reymond’s blocks work. Each block begins after the point where its integrand has dropped below all the earlier ones, and the construction never needs to know how far out that point is — only that it exists.

Series live on the same border #

For decreasing positive terms, a series and the corresponding integral converge or diverge together, and so the border between convergent and divergent series has exactly the same shape.

The integral test, due to Colin Maclaurin and Augustin-Louis Cauchy, compares each term $a_n = f(n)$ with the area under $f$ between $n$ and $n + 1$, and the difference between the sum and the integral stays bounded. So $\sum 1/(n \ln n)$ minus $\ln\ln N$ settles to a constant — the figure finds $0.795$ — and the sum diverges as slowly as its integral: ten million terms bring it only to $3.575$. Its partner $\sum 1/(n (\ln n)^2)$ converges, but its tail after $N$ terms is about $1/\ln N$, so after ten million terms it is still six hundredths short of its limit.

That is why summing a series numerically is no guide to whether it converges. Nor does rearranging help. The same terms, in a different order showed how rearranging a conditionally convergent series can make it add to anything — but that trick needs terms of both signs. For positive terms, every order gives the same sum, finite or infinite, so the border between convergent and divergent positive series is a property of the terms alone, and no clever arrangement moves a series across it. Joseph Bertrand’s family $\sum 1/(n (\ln n)^p)$, studied in 1842, converges exactly when $p > 1$; for $p$ near $1$ the partial sums of the convergent and divergent members are indistinguishable for longer than anyone could compute. Which functions can be added up met du Bois-Reymond on another boundary, where a continuous function’s Fourier series fails to converge; the tests that work there and here share the feature that no finite amount of computation settles them.

Why no test can be final #

The border’s absence has a practical consequence that is easy to state. Every convergence test compares the terms with a known scale — the ratio test with geometric series, the $p$-test with powers, Bertrand’s test with powers of logarithms, and further tests with iterated logarithms. Each test is decisive for series far enough from the border and inconclusive for series near it. Du Bois-Reymond’s theorem says that no scale is fine enough to decide every case: for any family of comparison integrands, there are integrands between them that the family cannot classify.

This is also why the series everything else is measured against could be the harmonic series only in a limited sense. It is the most famous divergent series and the natural first comparison, but it is not the slowest divergent series, and nothing is.

The substitution $x = e^u$ explains the self-similarity. It turns $\int dx/(x \ln x)$ into $\int du/u$, and $\int dx/(x \ln x \ln\ln x)$ into $\int du/(u \ln u)$. Each level of the chain is the previous level seen through one more logarithm, so a question about the $k$-th level is the same question one level down in a variable that is exponentially larger. The chain has no end for the same reason that exponentials can be stacked forever.

Hardy’s scale and orders of infinity #

Du Bois-Reymond went further than the theorem. He tried to build a calculus of rates of growth — an Infinitärcalcül — in which functions are compared by the eventual behaviour of their ratio, and the border problem was one of its first results.

G. H. Hardy tidied the subject in his 1910 tract Orders of Infinity. He showed that the functions built from $x$ by finitely many arithmetic operations, exponentials and logarithms — the logarithmico-exponential functions — are always eventually comparable: for any two of them, one eventually exceeds the other or their ratio tends to a limit. They form a totally ordered scale of growth rates, in which $\ln x$, $\ln\ln x$, $e^{\sqrt{\ln x}}$ and $x^{\varepsilon}$ each have a definite place. Du Bois-Reymond’s theorem says that no single function of that scale, or of any countable scale, marks the border of convergence.

An ordinal as a growth rate climbed a similar scale of growth rates in the opposite direction, towards ever faster functions, and met the same phenomenon: any countable list of rates can be outpaced by a diagonal construction. The slow direction and the fast direction are the same diagonal argument, run with the inequalities reversed.

What the figures can and cannot show #

The integrals are computed in closed form, and checked. Each running integral is an iterated logarithm, evaluated at numbers too large to write down by working with the number of digits instead of the number itself; one of them is checked against a direct numerical integration up to $10^{50}$. The tail step on the convergent side is checked the same way.

The diagonal construction is drawn for three blocks. The fourth block ends beyond any scale a figure can have; the construction’s claim — that it goes on forever and gains $1$ each time — is a matter of proof.

Divergence is never seen. Every divergent curve in these figures is a slowly rising line that stops at the edge of the plot. That the integral to infinity is infinite is a statement about what happens beyond every edge; the pictures show only that the curves have not yet levelled off, and on this question pictures deserve no trust at all.

Still open: whether a famous series converges #

The absence of a border is exactly what makes some individual series hard. A well-known example is the Flint Hills series,

$$\sum_{n=1}^{\infty} \frac{1}{n^3 \sin^2 n}.$$

Its terms are $1/n^3$ except when $n$ is very close to a multiple of $\pi$, where $\sin n$ is tiny and the term is enormous. Whether it converges depends on how well $\pi$ can be approximated by fractions — the question how close a fraction can get asked of every irrational number, measured here by the irrationality measure of $\pi$ — and Max Alekseyev showed in 2011 that convergence would imply that the irrationality measure is at most $5/2$. The best proved upper bound on that measure is about $7.1$, from Doron Zeilberger and Wadim Zudilin in 2020. Whether the series converges is unknown, and computing more terms cannot settle it: the terms that matter are the rare ones where $n$ lands extraordinarily near a multiple of $\pi$, and nobody can predict how near those will come.

A border made of nothing #

The habit worth keeping is to distrust a threshold found on one family.

For powers of $x$, the border between finite and infinite area sits exactly at $1/x$. It is tempting to believe that $1/x$ is the border for everything, and the belief fails immediately: $1/(x \ln x)$ is smaller and still diverges. Push further and the failure is total — below every divergent integrand is another, above every convergent one is another, and between the two families there is nothing. What looked like a line was a limit of two sequences approaching it from opposite sides, and du Bois-Reymond showed that the limit is not itself a function anyone can name.