Find the $1000$th digit after the decimal point of $\sqrt{n},$ where $n=\underbrace{11\dots1}_{1998 \text{ 1's}}$ 
Find the $1000$th digit after the decimal point of $\sqrt{n}$, where $n=\underbrace{11\dots1}_{1998 \text{ 1's}}$.

Obviously, $\underbrace{11\dots1}_{1998 \text{ 1's}}=\dfrac{1}{9}\left(9\cdot10^{1997}+9\cdot 10^{1996}+\dots+9\right),$ so we want to find $\left(\dfrac{10^{1998}-1}{9}\right).$ If only there was some way to convert this expansion into some closed form. I'm not sure if calculus would be useful. The problem asks for a single digit, so if we consider repeating digits, everything will be a lot easier. There seems to be a pattern in the decimal expansions of numbers consisting of only $1.$  For instance,
$$\sqrt{1}=1,$$
$$\sqrt{11}=3.3166247...$$
$$\sqrt{111}=10.5356537...$$
$$\sqrt{1111}=33.3316666...$$
$$\sqrt{11111}=105.408728...$$
$$\sqrt{111111}=333.333166...$$
$$\sqrt{1111111}=1054.09250...$$
Every term of the form $\displaystyle\sum_{n=0}^{2k+1}10^n$ has $k+1$ $3$'s at the beginning and $k+1$ 3's right after the decimal expansion, followed by one $1,$ and $2(k+1)$ $6$'s. Proving this would prove that the $1000$th digit is $1.$ This is the same as showing that $\sqrt{\left(\dfrac{10^{2m}-1}{9}\right)}=\dfrac{10^{m}-1}{3}+\dfrac{1}{3}-\dfrac{1}{6}\cdot 10^{-m}+\epsilon_m$ where $|\epsilon_m|<10^{-2m}.$
Edit: the previous question I asked was inspired by the current one, but the previous question seemed to have a rather unpleasant answer, so I changed it.

A little experimentation shows that if $n$ is the integer with $2m$ digits, where $m$ is an integer, all of them $1,$ then $\sqrt{n}$ has $m$ $3$'s, followed by a decimal point, $m$ $3$'s, a $1,$ and $2m$ $6$'s. If this was true, that would imply that the required digit is $1.$ Just substitute $2m=1998$ to verify.
Let $x=\dfrac{10^m-\frac{1}{2}10^{-m}}{3},m\in\mathbb{N}.$ Then $x$ has $m$ $3$'s, followed by a decimal point, followed by $m$ $3$'s, followed by one $1,$ followed by infinite $6$'s. So we just need to show that $(x-10^{-m-2})^2<n$ and $(x+10^{-m-2})^2>n,$ where $n=\dfrac{10^{2m}-1}{9}.$  This will show that the $(m+1)$st digit, or $1000$th digit, is indeed $1.$
$$(x-10^{-m-2})^2=\left(\dfrac{10^m}{3}-\dfrac{47}{300}10^{-m}\right)^2\\
=\dfrac{10^{2m}}{9}-\dfrac{47}{450}+\dfrac{2209}{90\;000}10^{-2m}\\
=n-\dfrac{47}{450}+\dfrac{2209}{90\;000}10^{-2m}\\
<n$$.
Similarly, 
$$(x+10^{-m-2})^2=\left(\dfrac{10^m}{3}+\dfrac{47}{300}10^{-m}\right)^2\\
=\dfrac{10^{2m}}{9}+\dfrac{47}{450}+\dfrac{2209}{90\;000}10^{-2m}\\
=n+\dfrac{47}{450}+\dfrac{2209}{90\;000}10^{-2m}\\
>n$$.
