Don't forget to carry

$S=\frac { 1 }{ 9 } +\frac { 1 }{ 99 } +\frac { 1 }{ 999 } +..\frac { 1 }{ { 10 }^{ n }-1 }=$

$0.11111111111111...+$

$0.01010101010101...+$

$0.00100100100100...$

First,it is clear that the $kth$ 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 $kth$ 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 since sums are preformed from right to left we must consider the possibility of a carry over from the $38$ th place.Since $38$ has four divisors $1,2,19$ and $38$ there are four $1$ 's in column $38$ . Of course,there is a chance that there is a carry over from the $39th$ place as well. However ,since the $38$ th place holds only four $1$ 's,it would take a carry of at least $6$ from the $39$ th place. We can show that the carry from the $39$ th place doesn't amount to $5$ . Thus we conclude there is no carry over and that the answer is indeed $2$ .

S= 9^-1 + 99^-1....
S/11= 99^-11+999^-1........
S-S/10=9^-1
then S=11/90 = 1.222222......

Rewrite the series

First rewrite the series into a more accessible form: $\begin{aligned} S&=\sum_{k=1}^\infty\frac{1}{10^k-1}=\sum_{k=1}^\infty \frac{10^{-k}}{1-10^{-k}}=\sum_{k=1}^\infty\sum_{l=1}^\infty (10^{-k})^l=\sum_{k=1}^\infty\sum_{l=1}^\infty10^{-kl}&\left|\text{geometric series, }|10^{-k}|\leq\frac{1}{10}<1\right. \end{aligned}$

The series converges absolutely for all $k,\:l\geq 1$ - we may reorder the series by the exponent $kl\overset{!}{=}d\in\mathbb{N}$ : $\begin{aligned} S&=\sum_{k=1}^\infty\sum_{l=1}^\infty10^{-kl}=\sum_{d=1}^\infty \sum_{(k,\:l)\in\Omega_d}10^{-nk}&\left|\Omega_d:=\{(k,\:l)\in\mathbb{N}^2|\quad nk=d\}\right.\\ \end{aligned}$

Notice that the inner sum is constant because the exponent equals $d$ ? In other words, the inner sum counts how many pairs exist such that $kl=d$ - that is the number of positive factors of $d$ : $\begin{aligned} S&= \sum_{d=1}^\infty c_d 10^{-d},&c_d=|\Omega_d|=\text{"number of positive factors of }d\text{"}\leq 2\sqrt{d}-1 \end{aligned}$

How many positive factors can $d=kl$ have? If it is not a perfect square, one factor has to be smaller than $\sqrt{d}$ and the other has to be greater than $\sqrt{d}$ - in total we have at most $2(\sqrt{d}-1)$ factors. If $d$ is a perfect square, we have one more factor $\sqrt{d}$ !

Converge back to the task

Let's find the $37^{th}$ digit of $S$ ! We may be tempted to say "that's $c_{37}$ , easy", but that might not be right: The coefficients $c_d$ are not digits and can be greater than 10. In other words, later coefficients might influence the $37^{th}$ digit as well! To make sure we get a correct estimate for the digit, we need to know how many terms we need to consider or how fast the series converges: $\begin{aligned} |c_d10^{-d}|&<2\sqrt{d}\cdot10^{-d}=:a_d,&&&\frac{a_{d+1}}{a_d}&=\frac{ 2\sqrt{d+1}\cdot 10^{-(d+1)} }{ 2\sqrt{d}\cdot 10^{-d} }=\sqrt{1+\frac{1}{d}}\cdot\frac{1}{10}\leq\frac{\sqrt{2}}{10}=:q<1\\[1em] \Rightarrow\quad |c_d10^{-d}|&<a_d\leq a_Nq^{d-N},&&&d&\geq N&(*) \end{aligned}$

Let's find out which digits the later coefficients of the series can influence: $\begin{aligned} \sum_{d=N}^\infty |c_d10^{-d}|&\underset{(*)}{<}\sum_{d=N}^\infty a_Nq^{d-N}=a_N\sum_{d=0}^\infty q^d\underset{|q|<1}{=}\frac{a_N}{1-q}=\frac{2}{1-0.1\sqrt{2}}\cdot \sqrt{N}\cdot10^{-N}\leq 2.4\cdot \sqrt{N}\cdot10^{-N}\\ \Rightarrow\quad \sum_{d=39}^\infty |c_d10^{-d}|&\leq 2.4\cdot\sqrt{39}\cdot10^{-39}<1.5\cdot 10^{-38} \end{aligned}$

We only need to consider two coefficients - $c_{37}=2$ and $c_{38}=4$ . Any following coefficients can at most flip the $38^{th}$ digit to $5$ and can be ignored - the $37^{th}$ digit of $S$ is indeed $\boxed{2}$ !

Rem.: For a challenge with overflow, prove that the $47^{th}$ digit of $S$ is $3\neq c_{47}$ !

37 can be divided by 1 and 37 so the first 1/9 and the eventual 1/10^37-1 will add a 1 in the 37th digit giving it a 2.

Here you can easily notice:

0.1111111111 1111111111 1111111111 1111111
0.0101010101 0101010101 0101010101 0101010
0.0010010010 0100100100 1001001001 0010010
0.0001000100 0100010001 0001000100 0100010
0.0000100001 0000100001 0000100001 0000100
0.0000010000 0100000100 0001000001 0000010
0.0000001000 0001000000 1000000100 0000100
0.0000000100 0000010000 0001000000 0100000
0.0000000010 0000000100 0000001000 0000010
0.0000000001 0000000001 0000000001 0000000
0.0000000000 1000000000 0100000000 0010000
0.0000000000 0100000000 0001000000 0000010
0.0000000000 0010000000 0000010000 0000000
0.0000000000 0001000000 0000000100 0000000
0.0000000000 0000100000 0000000001 0000000
0.0000000000 0000010000 0000000000 0100000
0.0000000000 0000001000 0000000000 0001000
0.0000000000 0000000100 0000000000 0000010
0.0000000000 0000000010 0000000000 0000000
0.0000000000 0000000001 0000000000 0000000
0.0000000000 0000000000 1000000000 0000000
0.0000000000 0000000000 0100000000 0000000
0.0000000000 0000000000 0010000000 0000000
0.0000000000 0000000000 0001000000 0000000
0.0000000000 0000000000 0000100000 0000000
0.0000000000 0000000000 0000010000 0000000
0.0000000000 0000000000 0000001000 0000000
0.0000000000 0000000000 0000000100 0000000
0.0000000000 0000000000 0000000010 0000000
0.0000000000 0000000000 0000000001 0000000
0.0000000000 0000000000 0000000000 1000000
0.0000000000 0000000000 0000000000 0100000
0.0000000000 0000000000 0000000000 0010000
0.0000000000 0000000000 0000000000 0001000
0.0000000000 0000000000 0000000000 0000100
0.0000000000 0000000000 0000000000 0000010
0.0000000000 0000000000 0000000000 0000001