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

Question

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.

Answer 1

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, $$(x+10^{-m-2})^2=\left(\dfrac{10^m}{3}+\dfrac{47}{300}10^{-m}\right)^2\\ =\dfrac{10^{2m}}{9}+\dfrac{47}{450}+\dfrac{2209}{90\;000}10^{-2m}\\ =n+\dfrac{47}{450}+\dfrac{2209}{90\;000}10^{-2m}\\ >n$$.

Answer 2

It seems you actually are on the right track. Solving "smaller" problems of the "same kind" often pays off, and this is one of the times it does.

You already found that

$$ n = \frac{10^{1998}-1}{9} = \frac{10^{1998}}{9} - \frac19.$$

The binomial expansion for $(a+b)^{1/2}$ with $a = \frac{10^{2m}-1}{9}$ and $b = -\frac19$ gives us \begin{multline} \left(\frac{10^{2m}}{9} - \frac19\right)^{\!1/2} = \left(\frac{10^{2m}}{9}\right)^{\!1/2} + \frac12 \left(\frac{10^{2m}}{9}\right)^{\!-1/2} \left(-\frac19\right)\\ - \frac18 \left(\frac{10^{2m}}{9}\right)^{\!-3/2} \left(-\frac19\right)^{\!2} + \frac1{16} \left(\frac{10^{2m}}{9}\right)^{\!-5/2} \left(-\frac19\right)^{\!3} + \cdots.\end{multline}

Now try the following comparisons: \begin{align} \left(\frac{10^{2m}}{9}\right)^{\!1/2} && \text{vs.} &&& \dfrac{10^{m}-1}{3}+\dfrac13, \\ \end{align} \begin{align} \frac12 \left(\frac{10^{2m}}{9}\right)^{\!-1/2} \left(-\frac19\right) && \text{vs.} &&& - \dfrac16 10^{-m} \\ \end{align} \begin{align} - \frac18 \left(\frac{10^{2m}}{9}\right)^{\!-3/2} \left(-\frac19\right)^{\!2} + \frac1{16} \left(\frac{10^{2m}}{9}\right)^{\!-5/2} \left(-\frac19\right)^{\!3} + \cdots && \text{vs.} &&& 10^{-2m} \end{align}

You should be able to confirm the formula that you worked out from the pattern of digits in $\sqrt{11},$ $\sqrt{1111},$ and $\sqrt{111111}.$