It's a well-known formula that $$ e = \lim\limits_{n\rightarrow\infty} (1+1/n)^n $$ and I'm wondering if there exists a similar formula for $\pi$. All the formulas I know for $\pi$ involve series or products or special functions (such as inverse trig functions). Or one can use Stirling's approximation if factorials are allowed. I'm wondering if there exists a function $F:\mathbb{Z}^m\rightarrow \mathbb{R}$ such that $F$ can be written in terms of a finite number of additions, subtractions, multiplications, divisions, and exponentiations, and $$ \pi = \lim\limits_{n\rightarrow\infty} F(\underbrace{n,n,\cdots,n}_{r \text{ times}},k_1,k_2,k_3,\dots,k_{m-r}) $$ for some fixed integers $m$, $r$, and $k_1,k_2,\dots,k_{m-r}$. In particular, a formula that is defined by a series, infinite product, recursive formula, or continued fraction would not satisfy these conditions.
To be perfectly clear about what I mean, let me formalize. Define a set of elementary sequences $S\subset \mathbb{R}^\mathbb{Z}$ as follows: Let $S_0=\{n\}\cup\{\text{all constant, integer-valued sequences}\}$ and define $S_i$ recursively by $$ S_i = \left\{ s_n+t_n,\, s_nt_n,\, (s_n)^{t_n},\, s_n-t_n,\, \frac{s_n}{t_n}\, \mid\, (s_n),(t_n)\in S_{i-1} \right\}. $$ Then let $S = \bigcup\limits_{i=0}^\infty S_i$. Does there exist a sequence $(s_n)\in S$ such that $\lim\limits_{n\rightarrow\infty} s_n =\pi$?
Note that $(s_n)\in S$ implies $(s_n)\in S_i$ for some finite $i$, which means that each term in the sequence $s_n$ has at most $i$ operations in it. So for example, the sequence $(1+1/n)^n$ is in $S_3$ because it involves 3 operations (one addition, one division, and one exponentiation). A series, such as $s_n = \sum\limits_{k=1}^n \frac{(-1)^k 4}{2k-1}$ is not in $S$ because the number of operations in the $n$th term is unbounded.
