2017th decimal place

Note that $N \; = \; \sum_{j=0}^{4031}10^j \; = \; \frac{10^{4032} - 1}{10 - 1} \; = \; \frac{10^{4032}}{9}\big[1 - 10^{-4032}\big]$ so that $\begin{aligned} \sqrt{N} & = \; \frac{10^{2016}}{3}\sqrt{1 - 10^{-4032}} \\ & = \; \frac{10^{2016}}{3}\big[1 - \tfrac12 \times 10^{-4032} + \cdots \big] \; = \; \frac13\big[10^{2016} - 5 \times 10^{-2017} + \cdots \big] \\ & = \; \frac13 \times \underbrace{999\ldots9}_{\times2015} \cdot \underbrace{999\ldots9}_{\times2016}50000.\ldots \\ & = \; \underbrace{333\ldots3}_{\times2015} \cdot \underbrace{333\ldots3}_{\times2016}16666.\ldots \end{aligned}$ and so the $2017$ th digit of $\sqrt{N}$ after the decimal point is $\boxed{1}$ . Note that the next term in the Binomial expansion of $\sqrt{1 - 10^{-4032}}$ is $-1.25 \times 10^{-8065}$ , so there will be a large number of $6$ s after that $1$ .