$n$-th digit in the sequence of natural numbers

The first step is to find what decade you are in. There are $9$ digits from the $1$ digit numbers, $2\cdot 90=180$ digits from the $2$ digit numbers for a total of $189$, and generally $n\cdot 9 \cdot 10^{n-1}$ from the $n$ digit numbers. Once you have found the decade, you can subtract the digits from the earlier decades. So if you want the $765$th digit, the first $189$ come from the first and second decades, so we want the $576$th digit of the $3$ digit numbers. This will come in the $\lceil \frac{576}3 \rceil=192$nd number, which is $291$. As $576 \equiv 3 \pmod 3$, the digit is $1$