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For any real number $r$ we can find a pair of natural numbers $N$ and $M$, such that $\sqrt{N}-\sqrt{M}$ will approximate $r$ with any given precision (if we choose $N,M$ large enough).

That's why I tried to think of any increasing sequences of integer numbers which would allow to compute $\pi$ in this way. And I mean, without knowing the digits of $\pi$ beforehand. Thus, $a_n, b_n$ have to be defined by recursion themselves. However, I wasn't able to make any progress.

1) The easiest way is to use the known value of $\pi$ to compute various pairs of $N,p$, such that:

$$\sqrt{N+p}-\sqrt{N} \approx \pi$$

$$N=\left[\frac{(p-\pi^2)^2}{4 \pi^2} \right]$$

Here $[]$ is the floor function. For example:

$$\sqrt{30268+1103}-\sqrt{30268}=3.14159$$

2) If we let the numbers be rational, and not necessarily increasing, we can use the known series to make something resembling what I want:

$$\pi^2=6 \left(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots \right)$$

$$\pi=\sqrt{6 \left(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots \right)}-\sqrt{0}$$


3) Another way, is to use the Spiral of Theodorus (spiral of square roots):

$$\rho (n)=\sqrt{n}$$

$$\phi (n)=\sum_{k=1}^{n} \arcsin \frac{1}{\sqrt{n}}-\frac{\pi}{2}$$

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The $\pi$ arises as a limit of radii difference at approximately the same angle:

$$\lim_{n \to \infty} \rho(\phi+2\pi)-\rho(\phi)=\pi$$

For example, the first such pair of numbers is:

$$\sqrt{21}-\sqrt{2}=3.17$$

However, I'm not sure if it's possible to properly define a sequence I want using this definition for a spiral - it's just too complicated (especially, finding pairs of rays with the closest angles).


Can we find sequences $\{a_n\}, \{b_n\}$ with $a_n,b_n \in \mathbb{Q}$, increasing and defined by recursion, such that $$\lim_{n \to \infty} (\sqrt{a_n}-\sqrt{b_n})=\pi$$?

Or can we prove that we can't?


Edit

I only consider rational sequences because I don't believe that any such integer sequences exist. However, I would be very happy if someone answers with integer sequences.

The main thing though, I want something different from $$\lim_{n \to \infty} \sqrt{a_n} = \pi+c, ~~~~ \lim_{n \to \infty} \sqrt{b_n}=c$$

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Let $\sum \limits _{k=0} ^\infty x_k$ be any series with positive and rational terms that converges to $\pi$. One such example may be

$$\sum _{k=0} ^\infty \frac 1 {16 ^k} \left( \frac 4 {8k+1} - \frac 2 {8k+4} - \frac 1 {8k+5} - \frac 1 {8k+6}\right)$$

but you may find others by considering MacLaurin series of various inverse trigonometric functions and evaluating them at special simple points (for instance the series of $\arcsin$ evaluated at $\frac 1 2$, multiplied by $6$).

Let

$$a_n = \left( \sum _{k=0} ^n x_k + 1 \right)^2, \quad b_n = \left( 1 - \frac 1 n \right) ^2 .$$

Notice that $\sqrt{a_n} - \sqrt{b_n} \to (\pi + 1) - 1 = \pi$. (The $1$ here is arbitrary, you may choose any other strictly positive number.)

To show that the sequences are recursive, notice that

$$a_{n+1} = a_n ^2 + 2 x_{n+1} a_n + x_{n+1} ^2, \quad b_{n+1} = b_n ^2 - \frac 2 {n (n+1)} b_n + \frac 1 {n^2 (n+1)^2} .$$

(Notice, though, that the recursion is not linear - but you have not asked for such a property.)

Finally, the sequences are increasing being squares of other increasing sequences (for $a_n$ here it is where the positivity of $x_n$ matters). It is also obvious that the sequences are made only of rational numbers (here is where the rationality of $x_n$ matters.)

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  • $\begingroup$ I knew someone would use a trick like this. I considered it myself. So, no 'interesting' sequences then. $\endgroup$ – Yuriy S Mar 20 '16 at 20:29
  • $\begingroup$ As for the recursion - I can be anything, not necessary linear. +1 for you, but I would like something not so trivial $\endgroup$ – Yuriy S Mar 20 '16 at 20:35
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I actually found a method to do what I want (without using a trivial $b_n$ sequence like in Alex' answer).

First we use a continued fraction for $\pi$ which has a regular structure, for example:

$$\frac{\pi}{4}=\cfrac{1}{1+\cfrac{1}{3+\cfrac{4}{5+\cfrac{9}{7+\cdots}}}}$$

Then (instead of taking partial convergents, like we usually do) we define a sequence of quadratic irrationals the following way:

$$x_1=\cfrac{1}{1+x_1}$$

$$x_2=\cfrac{1}{1+\cfrac{1}{3+x_2}}$$

$$x_3=\cfrac{1}{1+\cfrac{1}{3+\cfrac{4}{5+x_3}}}$$

And so on, we get a sequence of quadratic irrationals converging to $\pi$.

It's hard to define an explicit recurrence realtion for $x_n$ in this case, but I never said the recursion has to be explicit. The main point was to get a sequence of approximations in the form of root differences without knowing the digits of $\pi$ beforehand.

In this case we get:

$$4 x_1=2 \sqrt{5}-2=2.472\dots$$

$$4 x_2=2 \sqrt{21}-6=3.165\dots$$

$$4 x_3=\frac{\sqrt{745}}{2}-\frac{21}{2}=3.147\dots$$

$$4 x_4=\frac{\sqrt{49585}}{12}-\frac{185}{12}=3.140\dots$$

Both $a_n$ and $b_n$ are increasing (we can prove it) and defined by recursion (in a very broad sense).

Using other regular continued fractions for $\pi$ we can get other such approximations.

Using the simple continued fraction for $\pi$ we of course can get much better quadratic irrational approximations on the first few steps, such as:

$$3+\frac{\sqrt{57}}{2}-\frac{7}{2}=3.1400\dots$$

However, we have to know the digits of $\pi$ to get a simple continued fraction.

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  • $\begingroup$ How do these $x_n$ form a sequence given by recursion (even with a broader definition of recursion)? $\endgroup$ – Alex M. Mar 30 '16 at 10:12
  • $\begingroup$ In a very convoulted way. Since this continued fraction can be defined recursively, I'm sure it's possible to recover every $x_n$ from $x_{n-1}$ using the general form of the CF. I'm just not sure how. Anyway, I tried to clarify here what I meant in my question (to get an infinite sequence of approximations without knowing $\pi$). Note, that I will not change the question itself, so your answer is of course valid $\endgroup$ – Yuriy S Mar 30 '16 at 11:07
  • $\begingroup$ I aslo wanted both $a_n$ and $b_n$ unbounded, which I did not state in my question, so again, your answer is valid. Perhaps, I should delete this question and create a new one reflecting all the changes $\endgroup$ – Yuriy S Mar 30 '16 at 11:24
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$$a_{n+1} = a_n + \frac{24}{(n+1)^2} ,~~~ a_1 =25, ~~~a_2 =6 ,~~~ b_{n+1} = b_n + \frac{6}{(n+1)^2} , ~~~b_1=6 , ~~~b_2=\frac{6}{4}$$

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    $\begingroup$ You have a first order recurrences, why two initial conditions? They also do not appear to match each other and the relations $\endgroup$ – Yuriy S Mar 20 '16 at 18:42
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    $\begingroup$ I can't undertand what you mean by your expressions. Why $a_2$ and $b_2$ do not match the recurrence? If I use your relations with $a_1$ and $b_1$, I get $a_2=31$ and $b_2=15/2$ $\endgroup$ – Yuriy S Mar 20 '16 at 18:51
  • $\begingroup$ Anyway, I tried all four combinations of initial conditions, they don't work $\endgroup$ – Yuriy S Mar 20 '16 at 19:07

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