I'm new to this, forgive my formatting.

The task

Find the $1000$th decimal of $\sqrt{11...11}$ , where there are $1998$ $1$s.


I tried a smaller amount of $1$s:
$\sqrt{11} = 3.3166247903... =: \sqrt{a_1}$
$\sqrt{1111} = 33.3316666249... =: \sqrt{a_2}$
$\sqrt{111111} = 333.3331666666... =: \sqrt{a_3}$
There's a pattern, for an even amount of $1$s.
For $n$ $1$s, there will be $\frac{n}{2} 3$s before the decimal point and $\frac{n}{2} 3$s after it. Therefore when there are $1998$ $1$s, there will be:
$999$ $3$s . $999$ $3$s $1$.
In other words: $33...33.33....331$ Where the $1$ is the $1000$th decimal.

[EDIT] Forgot to mention that $i = \frac{n}{2}$. So we will get $1998$ $ 1$s if $i=999$.

Looking at the numbers that come after the chain of $3$s . I would want to prove that it will always be one.
$a_i = \sum_{k=0}^{2i-1} 10^k$ where $i\in\Bbb{N^+}$

To formulate this
$\sqrt{a_i} - \frac{3a_i}{10^i}$
Get the first number to the left of the decimal point, by multiplying it with $10$s.
$\biggl(\sqrt{a_i} - \frac{3a_i}{10^i}\biggl)10^{i+1}=: b_i$
$b_1=1.66247903..., b_2=1.6666249..., b_3=1.666666...,$ etc.
To prove that $1<b_i<2$, in other words, show that the series $b_i$ isn't decreasing.
For $<2:$
$\lim_{i\to \infty}b_i=\lim_{i\to \infty}\biggl(\sqrt{a_i} - \frac{3a_i}{10^i}\biggl)10^{i+1}=\frac{5}{3}=1.6666666666666666666666....$
The limit was calculated by Maple 2015.
The limit being $1.66666...$ means that $b_i$ will be getting closer to that value.
This proves the $<2$ part.


What I think I'm missing is proving that $b_i$ isn't decreasing.
Just because $b_1$ starts at $1.6624...$ and eventually $b_i$ will end up close to $1.666666666666...$ doesn't mean that it hasn't decreased somewhere along the way.
I tried showing that $b_i<b_{i+1}$ but got stuck at:

$\sqrt{a_i}<10\sqrt{a_{i+1}} -3 \cdot 10^i \cdot 11$
Maple is giving me a FAIL on the verify and evalb functions, which means that I don't know how to use Maple properly.
Maybe the fact that $a_i$ is rising can help somehow?

I may found a way, it's not nice.

We know that the first couple of values of $b_i$ are increasing, therefore I tried to show that $b_i$ is not increasing by finding the derivative of the function $b_i$ (thank you Maple).

Then having the derivative be equaled to zero, there is no such case, therefore there are no critical points, so the sequence doesn't change between increasing or decreasing, and since the first few values of $b_i$ were increasing, the entire sequence cannot be decreasing, making it non-decreasing.

Is this proof rock solid in general, or are there still some gaps to improve upon?

  • 4
    $\begingroup$ $$\sqrt{\frac{10^{1998}-1}{9}} = \frac{10^{999}}{3}\sqrt{1-\frac{1}{10^{1998}}}$$ and $\sqrt{1-x}$ is bounded between $1-\frac{x}{2}-\frac{x^2}{4}$ and $1-\frac{x}{2}$ for any $x>0$ close enough to the origin. Can you finish from here? $\endgroup$ – Jack D'Aurizio Nov 27 '17 at 20:14
  • $\begingroup$ @JackD'Aurizio shows, in particular, that your number (the square root) is less than $\frac{10^{999}}3 = 33...333.333333...... $ (the dot in the middle is the decimal point). On the other hand, show that your square root is greater than rounding that infinite decimal fraction to the first 1000 decimal places. $\endgroup$ – mathguy Nov 27 '17 at 20:30
  • $\begingroup$ @mathguy If I understand correctly, what you're saying is that $\sqrt{a_{999}} < \frac{10^{999}}{3}$ ? $\endgroup$ – Kappasenpai Nov 27 '17 at 21:01
  • $\begingroup$ @mathguy Could I show that by saying that $\sqrt{a_{999}} - \frac{3a_{999}}{10^i} = result$, therefore $\sqrt{a_{999}} = \frac{3a_{999}}{10^i} + result$ making $\sqrt{a_{999}} > \frac{10^{999}}{3}$? $\endgroup$ – Kappasenpai Nov 27 '17 at 21:05
  • $\begingroup$ @JackD'Aurizio I'm afraid not, I forgot to mention that $i=n/2$, meaning that there will be $1998$ $1$s if $i=999$, don't know if that changes anything. Either way, I'm not sure what you mean by " $\sqrt{1-x}$ is bounded between $1-\frac{x}{2}-\frac{x^2}{4}$ and $1-\frac{x}{2}$ for any $x>0$ close enough to the origin", could you please explain further? Thank you very much. $\endgroup$ – Kappasenpai Nov 27 '17 at 21:09

Your number is


By the Taylor expansion (or the generalized binomial theorem), this evaluates to


The $1000^{th}$ decimal is the rightmost digit of the integer part of (multiplying by $10^{1000}$)


which is $\lfloor\cdots333.333\cdots-1.666\cdots-\epsilon\rfloor=\cdots333\color{green}1$.

| cite | improve this answer | |
  • $\begingroup$ How can this $\frac{10^{999}}{3}\sqrt{1-10^{-1998}}$ be formulated? The Taylor expansion. Could you please explain in a little more detail, how it would work in this case? And how did you find $$\frac{10^{1999}}3-\frac{10}6-\frac1{24}10^{-1997}+\cdots$$ to be the $1000^{th}$ decimal, I understand how multiplying it by $10^{1000}$ helps, but I don't understand how it was picked out of the expansion. Also what does the $\epsilon$ represent in $\lfloor\cdots333.333\cdots-1.666\cdots-\epsilon\rfloor=\cdots333\color{green}1$. Thank you very much. $\endgroup$ – Kappasenpai Nov 28 '17 at 10:54
  • $\begingroup$ @Kappasenpai Pull the factor out of the radical. Lookup Taylor of $\sqrt{1+x}$. Write all terms as decimal numbers. $\epsilon$ is a neglectible quantity. $\endgroup$ – Yves Daoust Nov 28 '17 at 11:08
  • 1
    $\begingroup$ Oh, I see. The rest of the terms after the third power $-1997$ are so small, that we represent them with $\epsilon$ . $\endgroup$ – Kappasenpai Nov 28 '17 at 16:55

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