$ S = \frac19 + \frac{1}{99} + \frac{1}{999} + \dots + \frac{1}{10^n-1} + \dots \dots$

$ \Rightarrow S = 0.111111\dots + 0.010101\dots + 0.001001001\dots $

First, it is clear that the $k$ th column is composed only of $1$'s and $0$'s. The $1$'s in column $k$ occur in the $r$th row only if $r$ is a divisor of $k$. Below the $k$th row in column $k$ there are only $0$'s. Since $37$ is a prime, its only divisors are $1$ and $37$ and so the only $1$'s in column $37$ occur in rows $1$ and $37$. Thus we may be tempted to jump to the conclusion that the required digit is $2$. However, the sum can increase due to the carry over from column $38$, and the sum of column $38$ can increase due to the carry over from column $39$.