A sum whose terms vanish and whose total does not

There is a rule of thumb about infinite sums that almost everyone acquires without being taught it: if the terms shrink to nothing, the total settles down. Halves settle on 2. The reciprocals of the squares settle on $\pi^2/6$. It feels like the terms going to zero is what convergence is.

The harmonic series is the counterexample, and it is the first one worth meeting because it is so nearly the simplest series there is:

$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots$$

Its terms go to zero. Its total passes every number.

Oresme’s blocks #

The proof is about seven hundred years old, is three lines long, and is the reason this series is famous rather than merely awkward. Nicole Oresme gave it in the fourteenth century.

Group the terms in blocks whose lengths double: one term, then one, then two, then four, then eight.

$$1 \; + \; \frac{1}{2} \; + \; \underbrace{\frac13 + \frac14}_{} \; + \; \underbrace{\frac15 + \frac16 + \frac17 + \frac18}_{} \; + \; \underbrace{\frac19 + \cdots + \frac1{16}}_{} + \cdots$$

Now look at any one block. Every term in it is at least as large as the last term in it, and there are exactly enough terms to make that add up.

Take the block $\tfrac15 + \tfrac16 + \tfrac17 + \tfrac18$. Each of the four terms is at least $\tfrac18$, so the block is at least $4 \times \tfrac18 = \tfrac12$. Take the block from $\tfrac19$ to $\tfrac1{16}$: eight terms, each at least $\tfrac1{16}$, so at least $8 \times \tfrac1{16} = \tfrac12$.

Every block clears a half, and it is the same half every time, because doubling the number of terms exactly compensates for halving their size.

There are infinitely many blocks, each worth at least a half, so the total exceeds any bound: to pass $10$, take twenty blocks; to pass $50$, take a hundred. The series diverges, and the argument never computes a single sum.

What makes this proof good is what it refuses to do. It does not evaluate anything. It replaces every term by something smaller and shows that even the smaller thing is unbounded — which is a lower bound, and a lower bound is all divergence needs. Trying to add up the terms properly would have been much harder and would have answered a question nobody asked.

Divergent, and hopeless about it #

Divergence is a claim about eventually. It says nothing about how fast, and here the difference between the two is extreme enough to be worth its own figure.

The blocks argument gives the rate almost for free. To be sure of passing $n$ halves, take $2^n$ terms. So passing $10$ needs about a thousand terms, passing $20$ about a million, and passing $40$ about a million million. The total grows like the logarithm of the number of terms, which is the slowest growth that is still growth.

The offset is a constant with a name. The difference $H_n - \ln n$ settles on

$$\gamma = 0.5772156649\ldots,$$

the Euler–Mascheroni constant, which turns up throughout number theory and analysis and about which remarkably little is known — after two and a half centuries nobody has proved it irrational, let alone transcendental. It is one of the more embarrassing open questions in the subject: a number defined by a simple limit, computed to hundreds of billions of digits, and not known to be anything other than a fraction.

The appearance of a logarithm is not a coincidence either. The terms $1/n$ are the heights of the curve $1/x$, and adding them is stacking rectangles under that curve — whose area is a logarithm, since the logarithm is defined as that area in one of the standard treatments. The series and the integral differ by a bounded amount, and $\gamma$ is exactly the limit of that difference. Two of this collection’s earlier subjects meet here: the rectangles supply the growth, and the exponential’s inverse supplies the shape.

Practically, the growth rate means the divergence is unobservable. Summing a term per nanosecond since the universe began would reach a total somewhere around $70$. Reaching $100$ requires more terms than there are atoms in the observable universe. The series diverges and no computation will ever see it do so — the terms are the only visible part, and they all go to zero.

That is the reason this example matters more than its arithmetic suggests. It is a case where the finite evidence and the true answer point in opposite directions, and where no amount of extra finite evidence closes the gap. Watching the total is watching a logarithm, and a logarithm looks like it is stopping.

Leaning off a table #

The divergence can be built out of physical objects, which makes it harder to dismiss as an artefact of infinite processes.

Take identical blocks and stack them at the edge of a table, each shifted out from the one below. How far past the edge can the top block reach?

A single block balances at half its length. Two blocks: the top one overhangs the second by half, and the pair’s combined centre of mass sits a quarter along the second, so the second can overhang the table by a quarter. The pattern continues — the $k$-th block down can be pushed out by $1/(2k)$ of a block length — and the total overhang is

$$\frac12\left(1 + \frac12 + \frac13 + \cdots + \frac1n\right) = \frac{H_n}{2}.$$

Half the harmonic number, which is unbounded. A stack of ordinary blocks can lean out as far as anyone likes, with no glue, no counterweight, and no trick — the top block can be a mile past the table if there are enough below it.

And the logarithm makes the price plain. Reaching one block length of overhang needs four blocks. Two block lengths needs thirty-one. Three needs 227. Ten block lengths needs about $2.7 \times 10^8$ blocks, and each of them has to be placed to a tolerance that shrinks as fast as the overhang grows. The construction is unbounded and utterly impractical, in exactly the proportion the growth rate predicts.

The single-wide stack is not even optimal, which is a nice sting. Allowing several blocks per level — counterbalancing, as a bricklayer would — the achievable overhang grows like $n^{1/3}$ rather than $\ln n$, which is enormously faster. The harmonic stack is the answer to a specific question (one block per level) rather than to the general one, and it is the constrained version that produces the famous series.

Where it turns up without being sent for #

A series this slow is easy to file as a curiosity about infinity. It is not: $H_n$ is the answer to several ordinary counting questions, and in each of them the logarithm is the substantive part.

Collecting a set. Cereal packets contain one of $n$ equally likely cards. How many packets, on average, before the set is complete? Once $k$ cards are held, the chance a packet is useful is $(n-k)/n$, so the expected wait for the next new one is $n/(n-k)$. Summing over $k$ gives $n H_n$ — about $n \ln n$. For fifty cards that is $225$ packets rather than the fifty a first guess suggests, and the excess is entirely the harmonic tail: the last card alone costs an expected fifty packets.

Sorting. The average number of comparisons quicksort makes on random input is $2n H_n$ to within a constant, which is where the $n \log n$ in its reputation comes from. The logarithm is not an artefact of the analysis; it is the harmonic series, arriving because the chance that two particular elements are ever compared is $2/(j-i+1)$ and those reciprocals have to be summed.

Records. Read a random sequence of $n$ distinct numbers and count how often a new maximum appears. The $k$-th entry is a record exactly when it is the largest of the first $k$, which happens with probability $1/k$, so the expected number of records is $H_n$ — about $\ln n$. In a thousand readings expect about seven records; in a million, about fourteen. Doubling the data adds $0.7$ records, which is why record-breaking becomes rarer in a way that feels like the world settling down and is only arithmetic.

All three are the same sum wearing different clothes, and in all three the interesting behaviour is the logarithm rather than the divergence. That is the usual situation: the divergence is what makes the series famous, and the growth rate is what makes it useful — the same relationship a needle-dropping estimate has with its $1/\sqrt n$, where the convergence is guaranteed and the rate is the whole story.

Where the boundary sits #

The natural next question is what separates this series from the ones that do settle down, and the answer is sharper than “how fast the terms shrink”.

Reciprocals of squares converge. Reciprocals of $n^{1.01}$ converge. Reciprocals of $n$ do not. So the boundary sits exactly at the exponent $1$, and it is a genuine cliff rather than a gradual transition: every exponent above $1$ converges, the exponent $1$ diverges, and nothing in the terms’ behaviour signals which side of the line a given series is on.

Pushing further makes it stranger. The sum of $1/(n \ln n)$ also diverges — even more slowly, needing about $e^{e^{100}}$ terms to pass a hundred. The sum of $1/(n (\ln n)^{1.01})$ converges. There is an infinite hierarchy of series wedged between converging and diverging, each slower than the last, and no series among them is the slowest divergent one: given any divergent series of positive terms, a slower divergent series can always be constructed.

There is no boundary series. The convergent and divergent series of positive terms are not separated by a last case on either side, which is a fact worth holding on to whenever a classification looks like it ought to have a threshold in it.

Where it fails to diverge #

One more sting, and it is the one that shows how delicate the whole thing is.

Delete from the harmonic series every term whose denominator contains the digit $9$. What is left still contains almost all the small terms and looks barely thinner than the original.

It converges. Its total is under $23$.

The reason is a counting argument rather than an analytic one. Among the $d$-digit numbers there are $9 \times 10^{d-1}$ in total and only $8 \times 9^{d-1}$ with no $9$ in them, so the proportion surviving falls like $(9/10)^d$ — geometrically — while the terms only shrink like $10^{-d}$. Geometric decay beats the harmonic series, and the sum is finite. This is the Kempner series, and the same works for any digit and in any base.

So the divergence does not survive deleting a set of terms with density zero in the limit. That is worth sitting with next to Oresme’s proof, which makes divergence look robust and structural. It is structural, and it is also fragile: the blocks argument uses every term between $2^k$ and $2^{k+1}$, and a rule that removes even a thinning fraction of them can break it.

What the picture cannot show #

The bars in the opening figure stop at twenty-four and the claim is about infinitely many. That is the standing limitation of every drawing here, and in this case it is worse than usual, because the picture is actively suggesting the wrong conclusion: the bars are visibly heading for zero and the total is visibly slowing down, which is what a convergent series looks like at every finite stage.

This is the same trap Euler’s prime-generating polynomial sets from the other side: there, forty consecutive successes suggest a rule that fails at the forty-first; here, ten thousand terms of visible slowing suggest a limit that does not exist. In both cases the finite evidence is not a small sample of the infinite behaviour — it is a sample drawn from exactly the region where the behaviour has not started.

There is no number of terms at which a drawing of this series stops looking convergent. At ten thousand terms the running total is climbing by four ten-thousandths per step. At a million it is climbing by a millionth. The whole argument for divergence has to come from the blocking, which is a fact about how the terms are grouped and not about how they look.

The growth figure has the complementary problem. It shows the total tracking $\ln n$, which is honest, and $\ln n$ drawn on any axis a page can hold looks like it is levelling off. A logarithm is unbounded and no drawing of one is ever convincing about that, because the evidence for unboundedness is what happens past the right-hand edge.

The ladder from here #

Rungs above: the integral test, which compares the series to the area under $1/x$ and gives the logarithm directly — rectangles against a curve doing the work Oresme’s blocks do by hand. The $p$-series and where the cliff at $p = 1$ comes from. The alternating harmonic series, which converges to $\ln 2$, and the rearrangement theorem that lets it be rearranged to any total at all. Euler–Mascheroni in its own right. The Kempner series computed. The prime harmonic series, whose divergence is a stronger statement about the primes than their density alone. Optimal block stacking, and the $n^{1/3}$ result. And the connection to Fourier coefficients, where a square wave’s harmonics fall off exactly like $1/m$ and the Gibbs overshoot is what a $1/m$ tail costs.

The habit worth taking #

The useful thing here is not the series. It is the shape of the argument that settles it.

Oresme’s proof works by replacing every term with something smaller and simpler, and showing the replacement is already bad enough. Nothing is computed. No sum is evaluated. The bound is deliberately crude — the third block is worth $0.583$ and the argument only claims $0.5$ — and the crudeness is what makes it work, because a crude bound that is easy to compute infinitely often beats a sharp bound that cannot be.

That instinct transfers. When a quantity has to be shown unbounded, look for something smaller that is obviously unbounded. When it has to be shown finite, look for something larger that is obviously finite. Both moves throw away most of the problem, and the skill is in throwing away enough to make the remainder tractable without throwing away the answer — which is Euler deleting a city applied to arithmetic.