Finding the $1000$-th decimal of $\sqrt{1111...111}$

Question

As I was cleaning up my desk, I found my Calculus exam from almost a year ago. I remember there was only a bonus task that required either a tad more wit, either a bit more time. It goes like this :

---

$$ \text{Find the 1000-th decimal of }\underbrace{\sqrt{1111...111}}_{1998 \text{ times}} . $$

---

I remember noticing $11 = \frac{10^2-1}{9}$, building up a general case upon this observation and representing it via series using the binomial theorem, but nothing actually led me to the actual answer.

Any ideas ?

Answer 1

So you get $$ \frac13\sqrt{10^{1998}-1}=\frac13·10^{999}·\sqrt{1-10^{-1998}} $$ and by the binomial series $$ \sqrt{1+x}=1+\frac12x -\frac18x^2+\frac1{16}x^3-\frac{5}{128}x^4\pm… $$ it is only the first three terms that influence the first 4000 or so digits leading to \begin{align} \frac13·10^{999}·(1-5·10^{-1999}-1.25·10^{-3997}-…) &=\frac13·10^{999}·0.\underbrace{99…9}_{1998}4\underbrace{99…9}_{1997}874999… \\ &=10^{999}·0.\underbrace{33…3}_{1998}1\underbrace{66…6}_{1997}624999 \end{align} I'll leave you to find out where the 1000 decimal digit after the point comes to lie. To compare: \begin{align} \sqrt{999999}&=999.99949999987499993749996093747265622949217138670565794807… \\ \sqrt{111111}&=333.33316666662499997916665364582421874316405712890188598269… \\ \sqrt{99999999}&=9999.99994999999987499999937499999609374997265624979492187338… \\ \sqrt{11111111}&=3333.33331666666662499999979166666536458332421874993164062446… \end{align}