Could we calculate pi using an iterative series I know that, as a hobbyist mathematician, this is generally a term we can use to express pi
\begin{equation*} \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \frac{1}{13} - \frac{1}{15} + \frac{1}{17} - \cdots \end{equation*} 
This is great representation , and  it works just fine
However, I've been introduced to some iterative series recently (they're great for finding the roots to an equation by the way) and I was wondering, if there was a iterative series for pi, could we get a more justified value?
I've looked online, and there only seems to be geometric expressions using sine and cosine.
I was wondering if a numerical formula could be derived.
Edit: An iterative series is a series much like an algorithm, for example (N+1) = root(N+ 2/N) the idea being this series will converge on a value
 A: Well, there are iterative algorithms. There are two beautiful ones by the Borwein brothers, based on work by Ramanujan. Algorithm 1 involves the silver ratio, and Algorithm 2 involves the cube of the golden ratio.
I. Algorithm 1. Start with seed values:
$$y_0 = -1+\sqrt{2}$$
$$a_0 = 2(-1+\sqrt{2})^2$$
and two iterative rules,
$$y_{n+1} = \frac{1-(1-{y_n}^4)^{1/4}}{1+(1-{y_n}^4)^{1/4}}\tag1$$
$$a_{n+1} = a_n(y_{n+1}+1)^4-2^{2n+3}\,y_{n+1}\big(y_{n+1}^2+y_{n+1}+1\big)\tag2$$
Then,
$$\quad\quad\quad\lim_{n\to\infty} \frac{1}{a_n} = \pi\quad\text{(very fast)}$$
The difference grows quartically,
$$\quad\quad\quad\quad\frac{1}{a_n} - \pi \approx 4^{n+2} q^{4^n},\quad \text{where}\;q = e^{-2\pi}$$
Thus for $n=1,2,3$, the difference is about $10^{-10},\,10^{-42},\,10^{-172},$ or more than the fourth power of the previous. It's that fast.
A: Not sure what exactly it is that you want, but if you take newton's method and the power series of $\sin$ and $\cos$, you'll get $\pi$ as the limit of the newton iterations for $x_0 = 3$ and $x_{k+1} = x_k - \tan(x_k)$ where $\cot$ has a power series wich you can chose to evaluate up to an increasing degree, say $k$ to get
$$x_{k+1} = x_k - T_k[\tan](x_k)$$
Where $T_k[f]$ is the $k$-th sum of the Taylor series of $f$ at a predefined point near $\pi$ ($3$ for example). This will give you a new polynomial term each iteration but it forms a sequence with $\lim_{k\to\infty} x_k = \pi$.
See here for $T_k[\tan]$
A: The Basel infinite convergent serie (1/k squared) has the result of Pi squared over 6.By iterating actually you refine Pi value.There may be similar series that have a greater CONVERGING SPEED (you get PI decimals quicker).  
A: Here is a simple one:
Consider a sequence where $S_1=0$, $S_2=1$,
and $S_{n+2}=\begin{cases}
      \frac{(Sn+S_{n+1})}{2}, & \text{if $n$ is odd} \\
      \sqrt{Sn \cdot S_{n+1}}, & \text{if $n$ is even}
    \end{cases}$
Then $\lim\limits_{n\to\infty} \{\frac{2}{Sn}\}=\pi$.
