Number · the refusal to resolve
Cross Out Every Nine
The harmonic series adds forever. Delete every fraction whose bottom number contains a chosen digit — and the endless sum quietly stops.
Add up 1 + ½ + ⅓ + ¼ + ⅕ + …, one over every whole number, and the total never settles. It climbs past 10, past 100, past any number you name — slowly, like a coastline going out of sight, but it never stops. Nicole Oresme proved this around 1350, and it is the standard example of a sum whose terms shrink to nothing yet whose total is infinite.
Now cross out some of the terms. Not at random: strike out every fraction whose denominator, written out, contains the digit 9 — so 1/9 goes, and 1/19, 1/90, 1/1900, all of them. Add up only what's left. You would expect a slightly smaller infinity. Instead the sum converges — it lands on a finite number, 22.9206766…, and stays there. This is Kempner's series (1914), and it is not a rounding artifact. Pick a digit and watch it settle.
Delete every integer containing the digit…
Each cell is an integer, 1 upward. Lit cells survive; dark cells are crossed out for containing the digit.
A minority? No — almost everything.
The first reflex is that you have removed maybe a tenth of the numbers, so the sum should still be infinite, just thinner. But that intuition is calibrated on small numbers. Ask instead: among the numbers with exactly D digits, what fraction contain no 9 at all? Each digit dodges the 9 with probability at best 9/10, and the odds compound. By the time the numbers are long, avoiding the 9 everywhere is a near-miracle.
Among D-digit numbers, how many survive?
80.0% of 2-digit numbers contain no 9
8·9D−1 of the 9·10D−1 D-digit numbers survive.
So for large numbers you keep almost none of them — and the few you keep are enormous, with tiny reciprocals. Kempner's own argument makes the bound concrete: the survivors with D digits number fewer than 8·9D−1, each smaller than 1/10D−1, so their reciprocals add to less than 8·(9/10)D−1. Sum that over all lengths and the whole series is below 8·10 = 80 — finite, in one line, before you know its value. (The true value, 22.92, sits comfortably under the crude 80.)
You cannot get there by adding.
Knowing the sum is finite is not the same as finding it — and finding it by brute addition is hopeless. The survivors thin out so slowly that the running total barely moves. Set it going:
Add the survivors one at a time (no 9)
It stalls. Summing the reciprocals of all 43,046,720 no-9 integers below a hundred million — every single one — reaches only 13.2776, still more than nine short of the answer. The finite value instead comes from a different idea: group the survivors by how many digits they have, and carry from each length to the next not just the sum of reciprocals but the sums of reciprocal powers. Appending a digit then turns one length's totals into the next's, exactly. That is Baillie's method (1979) — and it is what the digit readout at the top of this page is running, live, in your browser.
Forbid a rarer pattern, and the sum grows.
Nothing above is special to single digits. Forbid any fixed string of digits — say the numbers must not contain 99 anywhere — and the same thing happens: convergence, to a finite value. But now the survivors are more plentiful (only the rare 99-containing numbers are gone), so the sum is larger. The rarer the forbidden string, the bigger the total — yet always finite, right up until the string is so rare you are keeping essentially everything, and the sum swells back toward the infinite harmonic series it came from.
Forbid any string of digits
Each extra 9 in the forbidden run multiplies the sum by roughly ten. The growth never stops — but for any fixed string, the total is finite.
The check
Every number on this page is recomputed two independent ways, and the engine you are running here is the one that was checked.
- The value. Deleting the digit 9 gives
22.92067661926415034816…— our recursion reproduces Baillie's published constant (Amer. Math. Monthly, 1979) to more than twenty digits. The ten single-digit sums increase in the orderno-1 < no-2 < … < no-9 < no-0: the commoner a digit, the more terms deleting it removes, the smaller the sum. - The independent witness. A completely separate program just
adds up survivors directly. Even 43,046,720 terms reach only
13.2776— a valid lower bound that sits below the recursion's value and climbs toward it, confirming the fast method from the outside. - This page's engine. The double-precision engine running in
your browser matches the 32-digit
mpmathreference to1e-10across all ten digits, and reproduces the gradient22.92 → 253.28 → 2555.87 → ≥25581.7. Verifier:research/kempner/— 8/8 checks pass.