Let $N$ be the positive integer with $2008$ decimal digits, all of them $1$. That is, $N=1111…1111$, with $2008$ occurrences of the digit $1$. Find the $1005th$ digit after the decimal point expansion of $\sqrt{N}$.
The proof given simply states two values and shows that their squares are greater than and less than $N$, and uses this to show that they are less than and greater than $\sqrt{N}$ and so the $1005th$ digit is $1$. This was from a calculator free exam so I don't see how this could have possibly been done, does anyone have any ideas?
Best Answer
Note that $N=\frac{10^{2008}-1}{9}$. One can use the Taylor series of $\sqrt{x}$ at $a=10^{2008}$ to solve this problem. More precisely, let $f(a)=\sqrt a$,
$$f(x)=f(a)+(x-a)f'(a)+\frac{f''(a)}2(x-a)^2+\cdots$$ Or $$\sqrt{9N}=10^{1004}-\frac{1}{2\cdot 10^{1004}}-\frac{3}{4\cdot 10^{3\cdot 1004}}+\cdots$$ That means $$3\sqrt{N}=10^{1004}-5\cdot 10^{-1005}-o(10^{-3000}).$$ The answer is obvious from here.