Where does the second '1' appear?

Question

Where does the second '1' appear in a value of following series?

$\frac{1}{9}+\frac{1}{99}+\frac{1}{999}+\dots$=$\sum_{n=1}^\infty {\frac{1}{10^n-1}}$

I already have a value of

Value of $x=\sum\limits_{n=1}^\infty {\frac{1}{10^n-1}} $ and location of second digit $1$ of $x$ (link)

But I want to know where the second '1' appears without this result.

There is easy way to determine poistion of the second '1'?

$$\frac{1}{9}+\frac{\color{red}{1}}{99}+\frac{1}{999}+\dots$$ — evil999man, May 10 '14 at 04:08
Why did you accept (25 minutes after the question was asked) an answer which does not (as far as I can see) answer the question? That is, except if a complete solution is behind the wall where Brilliant asks to register... — Did, Jul 25 '14 at 08:02

score 2 · Accepted Answer · answered May 10 '14 at 04:16

2

Hint: The decimal representation of $\frac{1}{10^n-1}$ has a one in every position that is a multiple of $n$.

For more details, read this problem. If you want more problems on this series, look at this and this.

answered May 10 '14 at 04:16

Calvin Lin

68,864

score 1 · Answer 2 · answered May 10 '14 at 04:37

Due to the ways the 1's are regularly repeated in the summation, the $n^{th}$ digit will, up to a certain point, equal the number of divisors of $n$. This will break down when the number of divisors exceeds 10.

Since every number > 1 has at least 2 divisors (itself and 1) we will not encounter another 1 until after the breakdown point.

A 1 will appear when we either reach a number with 10 divisors followed by a number with between 10 and 19 divisors.

From https://en.wikipedia.org/wiki/Table_of_divisors , 944 has 10 divisors, and 945 16 divisors, so the 944th digit will be a 1.

Where does the second '1' appear?

2 Answers2