Kempner’s Series

The harmonic series is 1 + 1/2 + 1/3 + 1/4 + …, the reciprocal of every whole number. Its terms shrink towards zero, which makes it tempting to think the sum must settle. It does not. Added far enough, the sum passes 10, then 100, then any number you name, although it needs 12,367 terms to pass 10 and a number of terms with more than 40 digits to pass 100.

Now cross out every term whose denominator, written in ordinary decimal digits, contains a 9. The terms 1/9, 1/19, 1/29 … 1/90 … 1/99 go, and so does 1/1990. The page lets you pick the digit, or a short string of digits, and shows the first hundred numbers with the survivors marked.

What is left is a series with infinitely many terms, and it does settle. Its sum is finite, about 22.92067. That is the surprise, and the rest of the page is about why and about how slowly it happens.

Among one-digit numbers, 1 of 9 contains a 9. Among two-digit numbers, 18 of 90 do. In every extra digit, another 9 can turn up, and the share of numbers that avoid it shrinks by a tenth each time: at six digits only about 53% of the numbers have no 9, and at twenty-four digits about 8%.

So the deleted terms are a small slice of the early terms and almost all of the later ones. The survivors in each block of numbers with the same number of digits grow ninefold from one block to the next, while the block itself grows tenfold, so their share keeps falling. The counting view tabulates this exactly.

The partial sums still climb, and very slowly. A million terms is not a small number, yet the sum of the surviving terms among them is about half of the limit. Nothing about a short run tells you where it ends, which is why the limit on this page is quoted from the literature rather than computed.

Sort the denominators by their number of digits. Numbers with k digits are all at least 10^(k−1), so their reciprocals are all at most 1/10^(k−1). Among those numbers, at most 8 · 9^(k−1) survive, because the first digit can be any of 8 choices (1 to 8) and every later digit any of 9 (0 to 8). The k-digit numbers therefore add at most 8 · 9^(k−1) / 10^(k−1) = 8 · (0.9)^(k−1).

That is a geometric series with ratio 0.9. It adds up to 8 · 10 = 80, whatever happens beyond. The true sum is much smaller, since the bound pretends that every surviving number is as small as it could be, which makes its reciprocal as large as possible, but a finite bound is all that convergence needs.

The harmonic series has no such saving: its k-digit block adds about the same amount, ln 10 ≈ 2.3, for every k, and adding 2.3 infinitely often is infinite.

Let $S$ be a string of $m$ decimal digits and let $A_k$ be the set of $k$-digit whole numbers whose decimal expansion does not contain $S$. Kempner's series for $S$ is

For a single non-zero digit,

For a string of $m$ digits, $|A_k|\le (10^m-1{)}^{\lfloor k/m\rfloor }10^{k\mathrm{mod}m}$ and

The harmonic series is the case with nothing deleted, and $\sum 1/n$ diverges.

The harmonic series first. For each $j\ge 0$, the $2^j$ terms from $1/(2^j+1)$ to $1/2^{j+1}$ are each at least $1/2^{j+1}$, so together they add at least $2^j/2^{j+1}=\frac{1}{2}$. There are infinitely many such groups, so the partial sums pass every bound.

Now delete every term containing a 9. Take the numbers with exactly $k$ digits and no 9. The first digit has 8 choices, 1 to 8, and each of the other $k-1$ digits has 9 choices, 0 to 8, so there are $8\cdot 9^{k-1}$ of them. Each is at least $10^{k-1}$, so each reciprocal is at most $10^{-(k-1)}$, and the whole group adds at most $8(9/10{)}^{k-1}$. Summing over every $k$,

Every partial sum of the series is at most the sum of the groups it touches, so every partial sum is below 80. The partial sums only grow, and an increasing sequence that is bounded converges. That finishes the proof for the digit 9, and it works for any non-zero digit. For 0 the first digit has 9 choices and the count is $9^k$, giving 90.

For a longer string $S$ of $m$ digits, cut the $k$ positions into $\lfloor k/m\rfloor$ blocks of $m$ consecutive digits, with $k\mathrm{mod}m$ positions left over. A number that avoids $S$ has no block equal to $S$, so each block has $10^m-1$ choices and the leftover positions have $10$ each. At most $(10^m-1{)}^{\lfloor k/m\rfloor }10^{k\mathrm{mod}m}$ numbers survive. Each reciprocal is again at most $10^{-(k-1)}$, so length $k$ adds at most $10(1-10^{-m}{)}^{\lfloor k/m\rfloor }$, and since each value of $\lfloor k/m\rfloor$ occurs for at most $m$ lengths, the total is at most $10m{\sum }_{j\ge 0}(1-10^{-m}{)}^j=10m10^m$. The bound is enormous, and finite.

What the proof does not give is the sum. The kernel checks the counts: it counts the survivors exactly with an automaton and compares with enumeration up to six digits, and checks that the counts never exceed the bound used here.

The partial sums on the crawl view are computed here, by adding doubles up to a million denominators, and the exact fractions on the first view are computed in exact arithmetic. The limits are not computed here. The sum for the digit 9, about 22.92067 66192 64150 34816, is Baillie's value (1979), and the value for the string 42, about 228.4463, is from Schmelzer and Baillie (2008), whose paper gives an algorithm for summing such slowly convergent series. Kempner proved convergence in 1914.

The page offers strings of up to three digits and denominators up to a million. A longer string converges too, by the same argument, and to a limit larger than any partial sum shown. The partial sums for 999 hardly differ from the full harmonic series at a million terms, and the limit of the 999-free series is far above anything the page could add up to.

The page does not say how large the limit is for every digit, and does not claim that partial sums approach the limit at any rate: the counting argument gives a bound, not a speed.