# “Complement” of Kempner Series

It is a long time since I summed any series. I was aware that the harmonic series diverged, (if I recall you can keep making groups that are greater than a half).

Then today I saw SMBC and it blew my mind. http://smbc-comics.com/index.php?id=3777 A quick google, and Wikipedia has an article on the Kempner series, although apparently it is normally nameless. https://en.wikipedia.org/wiki/Kempner_series This has an outline of the proof of convergence.

My question is, isn't the "Kempner" series, which is the harmonic series excluding numbers with the digit 9 in the denominator (in base 10), bigger than the series of numbers it excludes? (Surely it can't be smaller?) $$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}> \frac{1}{9}$$

But if you add them together you get infinity? Can you explain why? Is it just infinity messing with my intuition? A proof of the convergence or divergence of the sequence of omissions would be awesome.

Let $K^c$ be the sum of the reciprocals of all numbers with a $9$ in their decimal expansion. Consider the collection $T_n$ of numbers between $10^{n-1}$ and $10^n$ with a $9$. Then $|T_n|=10^n-8\cdot9^{n-1}$. Thus $$S_n=\sum_{a\in T_n}\frac1a\ge\frac{10^n-8\cdot9^{n-1}}{10^n}=1-\frac89\cdot\left(\frac{9}{10}\right)^n.$$ As $n\to\infty$, $S_n$ converges to something greater than or equal to $1$ and therefore $\sum S_n$ diverges.
The reason this happens is because most numbers are large and contain a nine. For instance, given a $100$-digit number at random, the probability that it doesn't contain a $9$ is approximately $\frac{8\cdot 9^{99}}{10^{100}}\approx.002\%.$ I hope this helps your intuition.