let
    S = 0.0
    k = 0
    tol = 1e-4 #you can change this into 1e-8 but it takes too long by design
    while true
        S += (-1.0)^k / (2k + 1)
        if abs(4S - pi) < tol
            break
        end
        k += 1
    end

    println("Smallest n = ", k)
    println("Approximation of π = ", 4S)
end

"""
Playing around with the tolerance (basically just lowering it and then gradually raising it, say start from tol=10^(-3), i.e. 1e-3, and dividing it by 10 each time) we find out that the number of iterations needed also multiply by 10, so for a tolerance of 1e-8 we need ~1 trillion iterations which is why this takes so long.

Part of programming is also questioning what one is already
"""