There are more numbers that contain digit 9 than numbers that do not contain digit 9

Remember from the last post that the sum of inverses of all strictly positive integers converges to infinity: 

S_1 = \displaystyle\lim_{n \to \infty}\displaystyle\sum_{i=1}^{n}\frac{1}{i} = \infty

I also proved that the sum of inverses of all strictly positive integers that don’t contain the digit 9 is finite. Denote this sum S_2.

Let S_3 be the sum of all strictly positive integers that contain the digit 9:

S_3=\frac{1}{ 9}+\frac{1}{ 19}+\frac{1}{ 29}+\cdots+\frac{1}{ 90}+\cdots+\frac{1}{ 99}+\frac{1}{ 109}+\cdots 

Then:

S_1 = S_2 + S_3

Thus S_3 = S_1 - S_2, and it follows that S_3=\infty because S_1=\infty and S_2 is finite.

In words: the sum of inverses of strictly positive integers that don’t contain digit 9 is finite and the sum of inverses of strictly positive integers that contain digit 9 is infinity.



This may be counterintuitive because apparently there are more numbers that don’t contain digit 9 than numbers that contain digit 9, but this is not actually true.

There are 9^n-1 strictly positive integers up to n digits that don’t contain digit 9 (see here). And obviously the total number of strictly positive integers is 10^n-1. This means that the number of strictly positive integers that contain digit 9 is: 10^n-9^n. And here’s the trick, for n large enough:  10^n-9^n>9^n-1.

Let’s plot 10^n-2 \cdot 9^n. You can see that for n=7 (horizontal axis) 10^n-9^n  is much bigger than 9^n-1.

10^n - 9^n > 9^n

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