An Interesting Question I Posed to Myself About $\pi$ as an Average. Prove or disprove:
There is a sequence $x$ with each $x_i\in\{1,2,3,4\}$ so that $\pi$ can be written as the average $$\pi = \lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{x_i}{n}$$
I am sure that this question would be trivial using advanced number theory concepts, but I would like a solution using just high-school olympiad level mathematics.
Thanks a lot. ☺
 A: Each term in either of these sums is equal to either $\left\lfloor\pi\right\rfloor = 3$ or $\left\lceil\pi\right\rceil = 4$:
\begin{align*}
\pi & = \lim_{n\to\infty}\frac{\left\lfloor{n\pi}\right\rfloor}{n} =
\lim_{n\to\infty}\frac1{n}\sum_{i=1}^n(\left\lfloor{i\pi}\right\rfloor -
\left\lfloor{(i - 1)\pi}\right\rfloor) \\
& = \lim_{n\to\infty}\frac{\left\lceil{n\pi}\right\rceil}{n} =
\lim_{n\to\infty}\frac1{n}\sum_{i=1}^n(\left\lceil{i\pi}\right\rceil -
\left\lceil{(i - 1)\pi}\right\rceil).
\end{align*}
A: You could define the sequence recursively in terms of the average of the previous terms of the sequence:
$$x_k = \begin{cases}
 3 & \text{ if } & a_{k-1}  > \pi \\
 4 & \text{ if } & a_{k-1} < \pi \\
\end{cases}$$
where
$$a_n = \frac {1}{n}\sum_k^n x_k$$

The convergence of $|a_n - \pi| \to 0$ follows from
$$- \frac{\pi - 3}{n} < a_n - \pi < \frac{4 - \pi}n$$
when $(x_{n-1}, x_n)$ is $(3, 4)$ or $(4, 3)$.  Also, $|a_n - \pi|$ is decreasing in the other cases.
In the $(3, 4)$ case, $a_{n-1} < \pi$ so $$\begin{array} {rcl} a_n &=& (a_{n-1}\cdot(n-1) + 4)/n \\ &<& (\pi \cdot (n-1) + 4)/n \\ &=& \pi + (4 - \pi)/n \end{array}$$
Similarly for the $(4, 3)$ case.

To be pedantically rigorous, it would also need to pointed out that there is no final time $a_n - \pi$ changes signs.
A: You can take
$n-\lfloor n(\pi-3)\rfloor$ times $3$ and $\lfloor n(\pi-3)\rfloor$ times $4$ and you get an error smaller than $\dfrac1n$, because the average is
$$3+\dfrac{\lfloor n(\pi-3)\rfloor}n=\dfrac{\lfloor n\pi\rfloor}n.$$
E.g. for $n=10000$, take $8585$ times $3$ and $1415$ times $4$, for average $3.1415$.

The digits $1$ and $2$ are more embarrassing than helpful.
A: Such a sequence clearly exists, for example I could say:
$$\frac{1}{7}(4+3+3+3+3+3+3)=\frac{22}{7}\approx\pi$$
and, continuing such a process towards an infinite number of terms, there exists a configuration that can get as close to $\pi$ (or any other number) as we like.
However, in terms of finding that configuration, I doubt there's much better than DanielV's answer.
A: Using only $3$s and $4$s, with $n$ of them you can make any integer number between $3n$ and $4n$.
Let $\frac{p}{q}$ be a convergent of the continued fraction of $\pi$: by choosing $n=q$ we may realize $p$ as a sum of $q$ numbers in $\{3,4\}$,
since $p>3q$ and $p<4q$. Moreover $\left|\pi-\frac{p}{q}\right|\leq \frac{1}{q^2}$. If we consider the concatenation of these sequences given by convergents we get an infinite sequence whose average value clearly converges to $\pi$ as wanted.
$$ \color{red}{\frac{3}{1}},\color{blue}{\frac{22}{7}},\color{purple}{\frac{333}{106}},\ldots\Longrightarrow \color{red}{3}\color{blue}{3333334}\color{purple}{3333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333444444444444444}\ldots $$
A: Yes.  Consider $a \le \omega \le b$. (In this specific case $a=3; b=4; \omega = \pi$)
Define $x_1=\begin{cases}b &\omega \le \frac {a+b}2\\a &\omega >\frac{a+b}2\end{cases}$
$v_k= average(x_1,....., x_k)=\frac {\sum_{i=1}^k x_i}k$.
$x_{k+1} = \begin{cases}b &\omega \le v_k\\a &\omega > v_k\end{cases}$.
It's easy to algebraically claim:

Claim 1:  $|v_{k+1} - v_k| \le \frac {b-a}{k+1}$

And it's easy to use that claim to claim by induction that

Claim 2: $|v_k - \omega| \le \frac {b-a}{k}$.

Then using the definition

Def:  $\lim_{n\to \infty}  v_n =\omega$ if when for any $\epsilon > 0$ there is an $N$ so that whenever $n > N$ then $|v_n -\omega| < \epsilon$.

the result follows:
Assuming $b > a$ (if $a=b$ then $\omega = a =b$ and $x_k = a_k = \omega = a=b$ and there is nothing to prove) then if we an $\epsilon > 0$ and we let $n > N \ge \frac 1{(b-a)}\epsilon$ then $|v_n - \omega| \le {b-a}{n+1} < \frac {b-a}n < \frac {b-a}N\le \epsilon$. So $\lim_{n\to \infty}  v_n =\omega$
A: Let me propose a "chemical approach" :
we want a mixture of  elements of atomic weight $\{ 1,2,3,4 \}$ such that the resulting average atomic weight is $\pi$.
We should then have the following diophantine system
$$ 
\left( {\matrix{   1 & 2 & 3 & 4  \cr    1 & 1 & 1 & 1  \cr  } } \right) 
\left( {\matrix{   {n_{\,1} }  \cr    {n_{\,2} }  \cr    {n_{\,3} }  \cr    {n_{\,4} }  \cr  } } \right) 
 = \left( {\matrix{   {22}  \cr    7  \cr  } } \right) 
$$
where the solutions shall be non-negative.
The system is under-determined so we are free to add some further bounds, for instance that the mixture be
somehow "centered", e.g.
$$ 
\eqalign{ 
  & \left( {\matrix{ 1 & 2 & 3 & 4 \cr  1 & 1 & 1 & 1 \cr 1 & { - 1} & { - 1} & 1  \cr 0 & { - 1} & 1 & 0  \cr  } } \right) 
\left( {\matrix{   {n_{\,1} }  \cr   {n_{\,2} }  \cr   {n_{\,3} }  \cr    {n_{\,4} }  \cr  } } \right) 
 = \left( {\matrix{   {N\,\pi }  \cr    N  \cr    0  \cr    0  \cr  } } \right)  \cr  
  & {\bf A}\;{\bf n} = N\;{\bf p} \cr}  
$$
Here I have chosen the matrix so that
$$ 
{\bf A}^{\, - \,1} \;{\bf p} = \left( {\matrix{ 
   {\left( {13 - 4\pi } \right)/12}  \cr  
   {1/4}  \cr  
   {1/4}  \cr  
   {\left( {4\pi  - 7} \right)/12}  \cr  
 } } \right) 
$$
the limit to which the ratio of the concentrations shall tend contains all positive values.
Then we can arrange the sequence in such a way that the proportion of the elements tend to the above.
So starting with $N=12$ we get
$$ 
{\bf n}_{\,12} 
  = \left( {\matrix{{13 - 4\pi }  \cr  3  \cr   3  \cr    {4\pi  - 7}  \cr  } } \right) 
\buildrel {\left\lfloor {} \right\rfloor } \over \longrightarrow 
 \left( {\matrix{   0  \cr    3  \cr    3  \cr   5  \cr  } } \right) 
\buildrel {n = 12} \over \longrightarrow 
 \left( {\matrix{   1  \cr    3  \cr    3  \cr    5  \cr  } } \right) 
$$
I took the floor, but rounding would be good as well, and optionally adjust for the total quantity.
The further step for e.g. $N=120$ gives
$$
{\bf n}_{\,120}  = \left( {\matrix{
   {\left( {13 - 4\pi } \right)10}  \cr 
   {30}  \cr 
   {30}  \cr 
   {\left( {4\pi  - 7} \right)10}  \cr 
 } } \right)\buildrel {\left\lfloor {} \right\rfloor } \over
 \longrightarrow \left( {\matrix{
   4  \cr 
   {30}  \cr 
   {30}  \cr 
   {55}  \cr 
 } } \right)\buildrel {n = N} \over
 \longrightarrow \left( {\matrix{
   5  \cr 
   {30}  \cr 
   {30}  \cr 
   {55}  \cr 
 } } \right)
$$
and we shall add to the sequence the ${\bf n}_{\,120} -{\bf n}_{\,12} $ elements.
