The sum that steps over every whole number

A sum whose terms vanish and whose total does not established that the harmonic sum $H(n) = 1 + \tfrac12 + \tfrac13 + \dots + \tfrac1n$ grows without bound — slowly, like the logarithm, but past every number eventually. So it passes 2, then 3, then 4, and every whole number after. The question that essay did not ask is whether it ever lands on one. It passes each whole number in steps of size $1/n$, and the steps get smaller and smaller; it is not obvious that one of them could not come down exactly on an integer.

None does. After $H(1) = 1$, no harmonic number is a whole number, and the proof takes a few lines once the right number is picked out of the list $1, 2, \dots, n$.

One term that nobody can cancel #

Count the factors of 2 in each of $1, 2, \dots, n$. The figure does it for $n = 12$: the odd numbers have none, $2$, $6$ and $10$ have one, $4$ and $12$ have two, and $8$ alone has three. That last fact is the whole proof, and it holds for every $n$. Among $1, \dots, n$ exactly one number carries the largest power of 2 — the largest power of two not exceeding $n$ itself. If two numbers both carried $2^m$, one of them would be at least $2 \cdot 2^m = 2^{m+1}$, and $2^m$ was chosen as the biggest power of two that fits.

Now put all the fractions over one denominator, the least common multiple $L$ of $1, \dots, n$ — the number whose prime powers are the largest each prime reaches in the range, which is the shape of a number’s divisors read the other way. It carries exactly $2^m$, since that is the highest power of two among the numbers. The term $1/k$ becomes $(L/k)/L$, and the numerator $L/k$ has the factors of 2 of $L$ minus those of $k$. For every $k$ except $2^m$ that leaves at least one factor of 2, so the numerator is even. For $k = 2^m$ it leaves none, so that numerator is odd.

Add them up: a sum of even numbers and one odd number is odd. So

$$H(n) = \frac{\text{odd}}{L},$$

with $L$ even as soon as $n \ge 2$. An odd number divided by an even one is never a whole number, and — because factorisation is unique — no rewriting of the fraction can move a factor of 2 from the denominator into the odd numerator. The proof is from 1915, due to Leopold Theisinger, and the only thing it uses about the terms is where the factors of two fall.

The table shows a little more than the proof needs. The power of 2 in the reduced denominator is not merely positive, it is exactly $2^m$ — it steps up at $n = 2, 4, 8, 16$ and stays level in between, because nothing else in the sum can reach the odd numerator the lonely term supplies. The rest of the denominator wanders about as primes enter and cancel; the power of two is the one part of it that is completely predictable.

A second proof, from a prime #

There is a proof that uses a prime instead of a power of two, and it is worth having because it shows the phenomenon is not special to the number 2.

For every $n \ge 2$ there is a prime $p$ with $n/2 < p \le n$. That is Bertrand’s postulate, proved by Chebyshev in 1852 and drawn in always one before the double. Such a prime divides exactly one of $1, \dots, n$ — itself — because its next multiple $2p$ is already past $n$. Over the common denominator, every numerator $L/k$ is then a multiple of $p$ except the one for $k = p$, so the total numerator is not divisible by $p$ while $L$ is. The prime stays in the denominator, and $H(n)$ is not whole.

Why the argument needs the prime 2 #

It is tempting to run the first proof with 3 in place of 2, and instructive that it fails. Among $1, \dots, 18$ the largest power of 3 that fits is $9$, and it is carried by two numbers, $9$ and $18$. Their terms are $\tfrac19 + \tfrac1{18} = \tfrac16$, and the two factors of 3 that each carried separately combine and cancel down to one. The reduced denominator of $H(18)$ is $4{,}084{,}080$, which is divisible by $3$ but not by $9$ — the power of 3 in the common denominator did not survive.

The power of 2 cannot be caught like this, and the reason is exact. Two multiples of $2^m$ that are not multiples of $2^{m+1}$ differ by at least $2 \cdot 2^m$, so both can fit below $n$ only if $2^{m+1}$ does — and then $2^{m+1}$ would be the largest power. For a prime $q$ the multiples of $q^m$ are only $q^m$ apart, and $q^m$ and $2q^m$ can both fit while $q^{m+1}$ does not. The number 2 is the only prime for which “the largest power that fits” automatically means “the only number carrying it”, because 2 is the only prime for which the next multiple is the next power.

That is also why the second proof reaches for a prime near the top of the range rather than a prime power. A prime $p$ above $n/2$ has only one multiple in range for the same reason $2^m$ does — its double is past $n$ — and Bertrand’s postulate is the guarantee that such a prime exists at every $n$.

The sum diverges twice #

There is a way of measuring numbers in which the power of 2 in a denominator is the size of the number. Declare a fraction large when its denominator carries many factors of 2, and small when its numerator does: the 2-adic size of $a/b$ is $2^{t}$, where $t$ is the number of factors of 2 in $b$ minus the number in $a$. Measured this way, $\tfrac1{1024}$ is enormous and $1024$ is tiny, and whole numbers are never larger than 1. It is the measure behind the parity patterns of the Collatz map, where it turned the map into something with a complete answer.

The table above is then a statement about 2-adic size. $H(n)$ has an odd numerator and a denominator carrying exactly $2^m$, so its 2-adic size is $2^m$, the largest power of two not exceeding $n$ — it grows without bound, just as the ordinary size does. The harmonic series diverges in two completely unrelated senses at once: in the usual one because the terms shrink too slowly, and in the 2-adic one because every few terms a new, larger power of two arrives in a denominator and nothing can cancel it. And the 2-adic divergence is the stronger proof that $H(n)$ is never whole, since a whole number has 2-adic size at most one.

The two proofs are the same idea wearing different clothes. Each finds a quantity — a power of two, or a prime — that occurs in exactly one term of the sum, so that term’s contribution cannot be matched by any other. Each needs a fact about how numbers are spread out: the powers of two double at each step, and Bertrand’s postulate says the primes are not too sparse. The first fact is trivial and the second is a theorem, which is why the power-of-two proof is the one usually given.

How close it comes #

A proof that $H(n)$ is never whole says nothing about how nearly whole it can be, and it can be quite near. Every time the sum passes a whole number, it does so by less than the last step it took.

The crossing points grow by a factor close to $e$ each time, because $H(n)$ is close to $\ln n + \gamma$, with $\gamma = 0.5772\ldots$ the constant the first essay met as the gap between the sum and the logarithm, and adding one to the logarithm multiplies $n$ by $e$. So the $k$-th crossing is near $e^{k - \gamma}$, and the step size there is near $e^{\gamma - k}$. The overshoots in the figure scatter below the line $1/n$ — sometimes close to it, sometimes far under — and at the crossing of 5 the sum lands just $0.0021$ past, at $n = 83$.

So the sum comes very close to whole numbers, closer and closer as they get larger, and the arithmetic of the denominator keeps it off every one of them. The two facts sit side by side without contradiction: a quantity can approach the integers as closely as desired and still never meet one, as long as something structural keeps the gap from being zero. A tail too small to be a whole number used the opposite version of the same device to prove $e$ irrational — a whole number trapped strictly between nought and one — and here the trap is an odd numerator over an even denominator.

Any run of consecutive reciprocals #

Nothing in the argument needed the sum to start at $1$. Take any run of two or more consecutive whole numbers, $m, m+1, \dots, n$. Among them, exactly one carries the most factors of 2. The reason is the same: if two did, say both divisible by $2^j$ and neither by $2^{j+1}$, they would be separated by an even multiple of $2^j$, and the number $2^j$ past the smaller of them — also in the run — would be divisible by $2^{j+1}$.

So the same proof applies to every such run, and the conclusion is Kürschák’s theorem of 1918: no sum of reciprocals of two or more consecutive whole numbers is a whole number. The figure computes all 190 sums with last term at most $1/20$ exactly, and each has an even denominator. The colours make visible a pattern the proof predicts: along a column the denominator’s power of two stays level while the run keeps containing the same dominant power of two, and changes only when the run’s start moves past it.

The consecutiveness matters. Reciprocals of distinct whole numbers can certainly add to a whole number — $\tfrac12 + \tfrac13 + \tfrac16 = 1$ — and that is the whole subject of Egyptian fractions. What rules it out here is that a consecutive run cannot avoid containing a single number with more factors of two than all its neighbours. Trygve Nagell and then Paul Erdős extended the theorem in the 1920s and 1930s to any arithmetic progression in place of consecutive numbers, where a prime plays the part of the lonely power of two.

The numerators know about squares #

If the denominators are this predictable, the numerators might be expected to be arbitrary. They are not, and one fact about them is genuinely surprising.

Take a prime $p$ and add the reciprocals of $1$ to $p - 1$. Pair the terms from the two ends: $\tfrac1k + \tfrac{1}{p-k} = \tfrac{p}{k(p-k)}$. Each pair is $p$ times something whose denominator is not divisible by $p$, so the numerator of $H(p - 1)$ is divisible by $p$. That much is a one-line observation. What is not obvious is that it is divisible by $p^2$.

That is Wolstenholme’s theorem, from 1862, valid for every prime from 5 on. The numerator of $H(4)$ is $25$, of $H(6)$ is $49$, of $H(10)$ is $7381 = 11^2 \cdot 61$, and so on, each an exact multiple of the square of the next prime. The second factor of $p$ comes from looking harder at the pairs. The sum of the pairs is $p$ times $\sum 1/(k(p-k))$, and modulo $p$ each $1/(k(p-k))$ is $-1/k^2$. As $k$ runs over the non-zero remainders, so does $1/k$, so $\sum 1/k^2$ is the same as $\sum k^2$, which is $\tfrac{(p-1)p(2p-1)}{6}$ — divisible by $p$ once $p$ is larger than $3$. That gives the second factor.

The table also shows the pattern stopping at two: the power of $p$ is exactly 2 in every row. A third factor would be a rare accident, and only two primes are known where it happens — $16{,}843$ and $2{,}124{,}679$, the Wolstenholme primes. The same theorem is often stated through binomial coefficients, where it says that $\binom{2p-1}{p-1}$ leaves remainder 1 on division by $p^3$, a statement much stronger than the remainder mod $p$ that the necklace proof of Fermat’s theorem would give.

Stated that way it invites the question every property of primes invites: does it pick out the primes? A composite $n$ for which $\binom{2n-1}{n-1}$ leaves remainder 1 on division by $n^3$ would pass for a prime by this test. None is known, searches have gone far without finding one, and whether one exists is open — the same situation as the power-sum test in every power sum, from the coefficients alone, except that there the first impostor turned up at 271,441 and here none has turned up at all. A congruence modulo $n^3$ is a much stricter sieve than one modulo $n$, and it may be strict enough to let nothing but primes through; nobody has a proof that it is.

What the fractions cannot show #

Every $n$. The tables stop at 16 and 20, the crossing plot at 4,550, the triangle of consecutive runs at 20. The theorems are proved for every $n$ by the counting of factors of two, and the figures are evidence that the counting was set up correctly, not a substitute for it. A reader looking only at the pictures would be entitled to suspect that some very large harmonic number might land on an integer; the argument about the unique power of two is what rules it out.

How the sum’s other primes behave. The power of two in the denominator is completely predictable, and the figures emphasise it. The rest of the denominator is not — primes enter it and can cancel out of it again, and the reduced denominator of $H(n)$ is sometimes much smaller than the least common multiple. Which primes cancel, and when, is a question about the numerators, and the pictures here only touch it.

The size of the numbers. The numerator of $H(31)$ has thirteen digits and that of $H(100)$ has forty. The tables print these in full while they fit; beyond that the exact fractions exist only in the computation, and no drawing conveys how quickly they outgrow a page.

Still open: which primes divide the numerators #

Wolstenholme’s theorem says $p^2$ divides the numerator of $H(p-1)$. A natural question is for which $n$ the prime $p$ divides the numerator of $H(n)$ at all. For $p = 3$ the answer is $n = 2$, $7$ and $22$ and nothing else; for $p = 5$ it is $n = 4$, $20$ and $24$. Both lists were found by computation and proved complete by an argument that works digit by digit in base $p$.

In 1991 Arulappah Eswarathasan and Eugene Levine conjectured that for every prime $p$ the list is finite. It is known to be finite for many small primes, by computations that follow the same digit-by-digit method, and for some of those primes the list is long. For primes in general, finiteness is not known, and the largest computations have found primes for which the search could not be completed. It is a question about the most familiar sum in mathematics, about which divisibility of its numerators, and nobody knows whether a single prime can divide infinitely many of them.

The lonely term #

The harmonic sum crosses every whole number and lands on none because, at every stage, one of its terms is alone: the reciprocal of the largest power of two, whose contribution to a common numerator has a parity no other term shares. A prime between $n/2$ and $n$ does the same job, and so does the dominant power of two in any run of consecutive numbers.

The same arithmetic, pointed at the numerators instead of the denominators, finds that the first $p - 1$ terms add up to a multiple of $p^2$ — a fact the pairing of terms explains only halfway, and the sum of squares modulo $p$ finishes. Whether the numerators keep avoiding a given prime forever after a while is open.