The aim of the kata is to try to show how difficult it can be to calculate decimals of an irrational number with a certain precision. We have chosen to get a few decimals of the number "pi" using
the following infinite series (Leibniz 1646–1716): 
PI / 4 = 1 - 1/3 + 1/5 - 1/7 + ... which gives an approximation of PI / 4.
http://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80
To have a measure of the difficulty we will count how many iterations are needed to calculate PI with a given precision. 
There are several ways to determine the precision of the calculus but to keep things easy we will calculate to within epsilon of your language Math::PI constant. In other words we will stop  the iterative process when the absolute value of the difference between our calculation and the Math::PI constant of the given language is less than epsilon.
Your function returns an array or an arrayList or a string or a tuple depending on the language (See sample tests) where your approximation of PI has 10 decimals 
In Haskell you can use the function "trunc10Dble" (see "Your solution"); in Clojure you can use the function "round" (see "Your solution");in OCaml or Rust the function "rnd10" (see "Your solution") in order to avoid discussions about the result.
Example :
```
your function calculates 1000 iterations and 3.140592653839794 but returns:
iter_pi(0.001) --> [1000, 3.1405926538]
```
Unfortunately, this series converges too slowly to be useful, 
as it takes over 300 terms to obtain a 2 decimal place precision. 
To obtain 100 decimal places of PI, it was calculated that
one would need to use at least 10^50 terms of this expansion!
About PI : http://www.geom.uiuc.edu/~huberty/math5337/groupe/expresspi.html
from math import pi

def iter_pi(epsilon):
    res = 0
    i = 0
    while abs(pi-res) > epsilon: 
        res += (-1)**(i%2) * 4 / (1 + (2*i))
        i += 1
    return [i, round(res, 10)]
from math import pi

def iter_pi(epsilon):
    res = 0
    i = 0
    while abs(pi-res) > epsilon: 
        res += (-1)**(i%2) * 4 / (1 + (2*i))
        i += 1
    return [i, round(res, 10)]